You are an environmental activist at heart but the reality is harsh and you are just a cashier in a cinema. But you can still do something!

You have n tickets to sell. The price of the i-th ticket is p_i. As a teller, you have a possibility to select the order in which the tickets will be sold (i.e. a permutation of the tickets). You know that the cinema participates in two ecological restoration programs applying them to the order you chose:

  * The x\% of the price of each the a-th sold ticket (a-th, 2a-th, 3a-th and so on) in the order you chose is aimed for research and spreading of renewable energy sources. 
  * The y\% of the price of each the b-th sold ticket (b-th, 2b-th, 3b-th and so on) in the order you chose is aimed for pollution abatement. 



If the ticket is in both programs then the (x + y) \% are used for environmental activities. Also, it's known that all prices are multiples of 100, so there is no need in any rounding.

For example, if you'd like to sell tickets with prices [400, 100, 300, 200] and the cinema pays 10\% of each 2-nd sold ticket and 20\% of each 3-rd sold ticket, then arranging them in order [100, 200, 300, 400] will lead to contribution equal to 100 ⋅ 0 + 200 ⋅ 0.1 + 300 ⋅ 0.2 + 400 ⋅ 0.1 = 120. But arranging them in order [100, 300, 400, 200] will lead to 100 ⋅ 0 + 300 ⋅ 0.1 + 400 ⋅ 0.2 + 200 ⋅ 0.1 = 130.

Nature can't wait, so you decided to change the order of tickets in such a way, so that the total contribution to programs will reach at least k in minimum number of sold tickets. Or say that it's impossible to do so. In other words, find the minimum number of tickets which are needed to be sold in order to earn at least k.

Input

The first line contains a single integer q (1 ≤ q ≤ 100) — the number of independent queries. Each query consists of 5 lines.

The first line of each query contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of tickets.

The second line contains n integers p_1, p_2, ..., p_n (100 ≤ p_i ≤ 10^9, p_i mod 100 = 0) — the corresponding prices of tickets.

The third line contains two integers x and a (1 ≤ x ≤ 100, x + y ≤ 100, 1 ≤ a ≤ n) — the parameters of the first program.

The fourth line contains two integers y and b (1 ≤ y ≤ 100, x + y ≤ 100, 1 ≤ b ≤ n) — the parameters of the second program.

The fifth line contains single integer k (1 ≤ k ≤ 10^{14}) — the required total contribution.

It's guaranteed that the total number of tickets per test doesn't exceed 2 ⋅ 10^5.

Output

Print q integers — one per query. 

For each query, print the minimum number of tickets you need to sell to make the total ecological contribution of at least k if you can sell tickets in any order.

If the total contribution can not be achieved selling all the tickets, print -1.

Example

Input


4
1
100
50 1
49 1
100
8
100 200 100 200 100 200 100 100
10 2
15 3
107
3
1000000000 1000000000 1000000000
50 1
50 1
3000000000
5
200 100 100 100 100
69 5
31 2
90


Output


-1
6
3
4

Note

In the first query the total contribution is equal to 50 + 49 = 99 < 100, so it's impossible to gather enough money.

In the second query you can rearrange tickets in a following way: [100, 100, 200, 200, 100, 200, 100, 100] and the total contribution from the first 6 tickets is equal to 100 ⋅ 0 + 100 ⋅ 0.1 + 200 ⋅ 0.15 + 200 ⋅ 0.1 + 100 ⋅ 0 + 200 ⋅ 0.25 = 10 + 30 + 20 + 50 = 110.

In the third query the full price of each ticket goes to the environmental activities.

In the fourth query you can rearrange tickets as [100, 200, 100, 100, 100] and the total contribution from the first 4 tickets is 100 ⋅ 0 + 200 ⋅ 0.31 + 100 ⋅ 0 + 100 ⋅ 0.31 = 62 + 31 = 93.






import java.io.*;
import java.util.*;
import java.math.BigInteger; 
import java.lang.*;





public class Main {
    
    
    
    
   static class sort implements Comparator<int[]>
   {
        public int compare(int[] a,int[] b)
        {
            
            return -b[0]+a[0];
        }
   }
   static class sort1 implements Comparator<long[]>
   {
        public int compare(long[] a,long[] b)
        {
            
          long c = -(a[0]-a[1]) + (b[0]-b[1]);

          if(c==0l) return 0;
          if(c>0l) return 1;
          return -1;
        }
   }
   static class sort5 implements Comparator<long[]>
   {
        public int compare(long[] a,long[] b)
        {
            
           double d1,d2;
           d1 = (a[0]+0.00)/(a[1]+0.00);
           d2 = (b[0]+0.00)/(b[1]+0.00);
           //int v = Double.comapre(d2,d1);
           double eph = 0.000000000000001;
           double d = (d2-d1);
          // pr(d1+" "+d2);
           if(d>eph) return 1;
           else if(d<eph) return -1;
           if(a[1]>b[1]) return 1;
           else if(b[1]>a[1]) return -1;
           return 0;
        }
   }
   static class sort11 implements Comparator<double[]>
   {
        public int compare(double[] a,double[] b)
        {
            //if(a[0] == b[0]) return a[1]-b[1];
           if(a[1] < b[1]) return -1;
           else if(a[1] > b[1]) return 1;
           return 0;
        }
   }
   public static String[] F(BufferedReader bf) throws Exception
    {
        return (bf.readLine().split(" "));
    }
    
   public static void pr(PrintWriter out,Object o)
   {
      
    out.println(o.toString());//out.flush();
    
   }
   public static void prW(PrintWriter out,Object o)
   {
      
    out.print(o.toString());//out.flush();
    
   }
   public static int intIn(String st)
   {
      return Integer.parseInt(st);
   }
   
   
    
    public static void pr(Object o)
    {
        System.out.println(o.toString());
    }
    public static void prW(Object o)
    {
        System.out.print(o.toString());
    }
    
   
    public static int inInt(String s)
    {
        return Integer.parseInt(s);
    }
    public static long in(String s)
    {
        return Long.parseLong(s);
    }
 
 
    static int[] toIntArray(String[] m) 
    { 
        int[] p=new int[m.length];
        for(int o=0;o<m.length;o++)
        {
            p[o]= inInt(m[o]);
        }
        return p;
    }
    static double[] toDArray(String[] m) 
    { 
        double[] p=new double[m.length];
        for(int o=0;o<m.length;o++)
        {
            p[o]= Double.parseDouble(m[o]);
        }
        return p;
    }
    static long[] toLArray(String[] m) 
    { 
       long[] p=new long[m.length];
        for(int o=0;o<m.length;o++)
        {
            p[o]= in(m[o]);
        }
        return p;
    }
   
     static long gcd(long a, long b)
        {
          if (b == 0)
            return a;
          return gcd(b, a % b); 
        }
     
        static long pow(long x, long y, long p)
      {
        if(y == 0) return 1l;
        long res = 1; // Initialize result
     
        x = x % p; // Update x if it is more than or
        // equal to p
     
        if (x == 0)
          return 0l; // In case x is divisible by p;
     
        while (y > 0)
        {
     
          // If y is odd, multiply x with result
          if ((y & 1) != 0)
            res = (res * x) % p;
     
          // y must be even now
          y = y >> 1; // y = y/2
          x = (x * x) % p;
        }
        return res;
      }
    
   
    static int nCr(int n, int r, int p)
    {
        if (r > n - r)
            r = n - r;
 
        // The array C is going to store last
        // row of pascal triangle at the end.
        // And last entry of last row is nCr
        int C[] = new int[r + 1];
 
        C[0] = 1; // Top row of Pascal Triangle
 
        // One by constructs remaining rows of Pascal
        // Triangle from top to bottom
        for (int i = 1; i <= n; i++) {
 
            // Fill entries of current row using previous
            // row values
            for (int j = Math.min(i, r); j > 0; j--)
 
                // nCj = (n-1)Cj + (n-1)C(j-1);
                C[j] = (C[j] + C[j - 1]) % p;
        }
        return C[r];
    }
    // F(o,-1,tm,vis,par,o,mat);
   public static int log(int x)
   {
       boolean bol = true;

       if(((x-1)&x) == 0) bol=false;

       int c=0;
       while(x > 0)
       {
        x=(x/2);
        c++;
       }

       if(!bol) c--;
       return c;

   }

   
   static class sor implements Comparator<long[]>
   {
      public int compare(long[] a,long[] b)
      {
         if(a[0] == b[0])
         {
            if(a[1] < b[1]) return -1;
            return 1;
         }
         if(a[0] > b[0]) return 1;
         return -1;
      }
   }


    static long ncr(int n, int r)
    {
 
        // p holds the value of n*(n-1)*(n-2)...,
        // k holds the value of r*(r-1)...
        long p = 1, k = 1;
        if(n<r) return 0l;
 
        // C(n, r) == C(n, n-r),
        // choosing the smaller value
        if (n - r < r) {
            r = n - r;
        }
 
        if (r != 0) {
            while (r > 0) {
                p *= n;
                k *= r;
 
                // gcd of p, k
                long m = __gcd(p, k);
 
                // dividing by gcd, to simplify
                // product division by their gcd
                // saves from the overflow
                p /= m;
                k /= m;
 
                n--;
                r--;
            }
 
            // k should be simplified to 1
            // as C(n, r) is a natural number
            // (denominator should be 1 ) .
        }
        else {
            p = 1;
        }
 
        // if our approach is correct p = ans and k =1
        return p;
    }
 
    static long __gcd(long n1, long n2)
    {

        if(n1==0l) return n2;
        if(n2==0l) return n1;
        if(n1==1l || n2==1l) return 1l;
       // long gcd = 1;
        if(n1 == n2)  return n1;
        if(n1>n2) return __gcd(n1%n2,n2);
        return __gcd(n1,n2%n1);
        
    }
  
    
    public static void main (String[] args) throws Exception {
        
        BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);;;
        
        //int[] map=new int[1000001];
      
       int yy=inInt(bf.readLine());
       


        
        for(int w=0;w<yy;w++)
        {
            // String str = bf.readLine();
          out.flush();
         
            String[] xlp = bf.readLine().split(" ");;;;;;
            //String st = bf.readLine();
         int n;//boolean bol=false;
           long m;//long a,b,c;
           int ioo;
           long  k;//pr(out,"vbc");
         n=inInt(xlp[0]);//m=in(xlp[1]);//k=in(xlp[2]);//t=inInt(xlp[3]);

         long[] arr = toLArray(F(bf)); 

         int[] tem = toIntArray(F(bf));

         int xc,a,yc,b;
         xc = tem[0];a=tem[1];
         tem = toIntArray(F(bf));

         
         yc = tem[0];b=tem[1];

         k = in(bf.readLine());
         long x,y;x=xc+0l;y=yc+0l;
         Arrays.sort(arr);

         int lo,hi;lo=0;hi=n;
         long ans = n+1;
         long lcm = ((a+0l)*(b+0l))/((__gcd(a,b)));
         while(lo < hi)
         {
            int mid = (lo+hi)/2;

            int i=n-1;
            long res=0l;
            int c=0;
                  for(long j=lcm;j<=mid;j+=lcm)
                  {
                       // if(c>mid) break;
                        res += ((arr[i]*(x+y))/(100l));
                        //if((j%b) == 0) res += ((arr[i]*y)/(100l));
                        i--;
                        c++;
                  }
            if(x>y)
            {
                  for(int j=a;j<=mid;j+=a)
                  {
                       // if(c>mid) break;
                        if((j%lcm) != 0){
                        res += ((arr[i]*x)/(100l));
                        i--;
                        c++;}
                  }
                 // i=n-1;
                   for(int j=b;j<=mid;j+=b)
                  {
                        //if(c>mid) break;
                        if((j%lcm) != 0){
                        res += ((arr[i]*y)/(100l));
                        i--;
                        c++;}
                  }
            }
            else
            {
                  for(int j=b;j<=mid;j+=b)
                  {
                        //if(c>mid) break;
                        if((j%lcm) != 0){
                        res += ((arr[i]*y)/(100l));
                        i--;
                        c++;}
                  }
               //   i=n-1;
                  for(int j=a;j<=mid;j+=a)
                  {
                        //if(c>mid) break;
                        if((j%lcm) != 0){
                        res += ((arr[i]*x)/(100l));
                        i--;
                        c++;}
                  }
                   
            }

            if(res >= k)
            {
                  ans = mid;
                  hi=mid-1;
            }
            else lo=mid+1;

         }

         int mid = lo;

            int i=n-1;
            long res=0l;
            int c=1;
            for(long j=lcm;j<=mid;j+=lcm)
                  {
                       // if(c>mid) break;
                        res += ((arr[i]*(x+y))/(100l));
                        //if((j%b) == 0) res += ((arr[i]*y)/(100l));
                        i--;
                        c++;
                  }
            if(x>y)
            {
                  for(int j=a;j<=mid;j+=a)
                  {
                       // if(c>mid) break;
                        if((j%lcm) != 0){
                        res += ((arr[i]*x)/(100l));
                        i--;
                        c++;}
                  }
                 // i=n-1;
                   for(int j=b;j<=mid;j+=b)
                  {
                        //if(c>mid) break;
                        if((j%lcm) != 0){
                        res += ((arr[i]*y)/(100l));
                        i--;
                        c++;}
                  }
            }
            else
            {
                  for(int j=b;j<=mid;j+=b)
                  {
                        //if(c>mid) break;
                        if((j%lcm) != 0){
                        res += ((arr[i]*y)/(100l));
                        i--;
                        c++;}
                  }
               //   i=n-1;
                  for(int j=a;j<=mid;j+=a)
                  {
                        //if(c>mid) break;
                        if((j%lcm) != 0){
                        res += ((arr[i]*x)/(100l));
                        i--;
                        c++;}
                  }
                   
            }
            if(res >= k)
            {
                  ans = Math.min(ans,mid);
                  //hi=mid-1;
            }

            if(ans > n) pr(out,-1);
            else pr(out,ans);




          
          




      

         
         
         



         
         
   }
    out.close();
   bf.close();
}}


/*
2
1999999973 1999999943
1999999973 1999999943

Kickstart
 String rp;
            rp = "Case #"+(w+1)+": "+(n-ans)+" ";
static int[][] dir={{0,1},{1,0},{-1,0},{0,-1}};

static class SegmentTreeRMQ 
    { 
    int st[]; 
    int minVal(int x, int y) { 
        return (x > y) ? x : y; 
    } 
  
    
    int getMid(int s, int e) { 
        return s + (e - s) / 2; 
    } 
  
    
    int RMQUtil(int ss, int se, int qs, int qe, int index) 
    { 
        
        if (qs <= ss && qe >= se) 
            return st[index]; 
  
        // If segment of this node is outside the given range 
        if (se < qs || ss > qe) 
            return Integer.MIN_VALUE; 
  
        // If a part of this segment overlaps with the given range 
        int mid = getMid(ss, se); 
        return minVal(RMQUtil(ss, mid, qs, qe, 2 * index + 1), 
                RMQUtil(mid + 1, se, qs, qe, 2 * index + 2)); 
    } 
  
    // Return minimum of elements in range from index qs (query 
    // start) to qe (query end).  It mainly uses RMQUtil() 
    int RMQ(int n, int qs, int qe) 
    { 
        // Check for erroneous input values 
        
  
        return RMQUtil(0, n - 1, qs, qe, 0); 
    } 
  
    // A recursive function that constructs Segment Tree for 
    // array[ss..se]. si is index of current node in segment tree st 
    int constructSTUtil(int arr[], int ss, int se, int si) 
    { 
        // If there is one element in array, store it in current 
        //  node of segment tree and return 
        if (ss == se) { 
            st[si] = arr[ss]; 
            return arr[ss]; 
        } 
  
        // If there are more than one elements, then recur for left and 
        // right subtrees and store the minimum of two values in this node 
        int mid = getMid(ss, se); 
        st[si] = minVal(constructSTUtil(arr, ss, mid, si * 2 + 1), 
                constructSTUtil(arr, mid + 1, se, si * 2 + 2)); 
        return st[si]; 
    } 
  
    
    void con(int arr[]) 
    { 
        // Allocate memory for segment tree 
  
        //Height of segment tree 
        int n = (arr.length);
        int x = (int) (Math.ceil(Math.log(n) / Math.log(2))); 
  
        //Maximum size of segment tree 
        int max_size = 2 * (int) Math.pow(2, x) - 1; 
        st = new int[max_size]; // allocate memory 
  
        // Fill the allocated memory st 
        constructSTUtil(arr, 0, n - 1, 0); 
    }
    }
     static class DSU {
    
    int[] p;int[] sz;int op;int c;;
    int[] last;
    public void G(int n)
    {
        last=new int[n];
        p=new int[n];
        sz=new int[n];c=n;
        op=n;
        for(int h=0;h<n;h++)
        {
            sz[h]=1;p[h]=h;
            last[h]=h;
        }
    }
    public int find(int x)
    {
        int y=x;
        while(x!=p[x]) x=p[x];
        while(y!=p[y])
        {
            int tem=p[y];
            p[y]=x;y=tem;
        }
        return p[y];
    }
    public void union(int a,int b)
    {
        int x,y;
        x=find(a);y=find(b);
        if(x==y) return;
        if(sz[x]>sz[y])
        {
            p[y] = x;
            sz[x]+=sz[y];
            last[x]=Math.max(last[x],last[y]);
        }
        else
        {
            p[x]=y;sz[y]+=sz[x];
            last[y]=Math.max(last[y],last[x]);
        }
        c--;
        
    }}

    static long pow(long x, long y, long p)
      {
        long res = 1; // Initialize result
     
        x = x % p; // Update x if it is more than or
        // equal to p
     
        if (x == 0)
          return 0l; // In case x is divisible by p;
     
        while (y > 0)
        {
     
          // If y is odd, multiply x with result
          if ((y & 1) != 0)
            res = (res * x) % p;
     
          // y must be even now
          y = y >> 1; // y = y/2
          x = (x * x) % p;
        }
        return res;
      }
      static long gcd(long a, long b)
        {
          if (b == 0)
            return a;
          return gcd(b, a % b); 
        }
        static int gcd(int a, int b,int o)
        {
          if (b == 0)
            return a;
          return gcd(b, a % b,o); 
        }

Geometric median

public static double F(double[] x,double[] w)
    {
        double d1,d2;
        double S=0.00;
        for(double dp : w) S += dp;
        int k = 0;
        double sum = S - w[0]; // sum is the total weight of all `x[i] > x[k]`

        while(sum > S/2)
        {
            ++k;
            sum -= w[k];
        }
        d1=x[k];
        return d1;
        k = w.length-1;
       sum = S - w[k]; // sum is the total weight of all `x[i] > x[k]`

        while(sum > S/2)
        {
            --k;
            sum -= w[k];
        }
        d2=x[k];
        return new double[]{d1,d2};
    }
*/



























