You play a computer game. In this game, you lead a party of m heroes, and you have to clear a dungeon with n monsters. Each monster is characterized by its power a_i. Each hero is characterized by his power p_i and endurance s_i.

The heroes clear the dungeon day by day. In the beginning of each day, you choose a hero (exactly one) who is going to enter the dungeon this day.

When the hero enters the dungeon, he is challenged by the first monster which was not defeated during the previous days (so, if the heroes have already defeated k monsters, the hero fights with the monster k + 1). When the hero fights the monster, there are two possible outcomes:

  * if the monster's power is strictly greater than the hero's power, the hero retreats from the dungeon. The current day ends; 
  * otherwise, the monster is defeated. 



After defeating a monster, the hero either continues fighting with the next monster or leaves the dungeon. He leaves the dungeon either if he has already defeated the number of monsters equal to his endurance during this day (so, the i-th hero cannot defeat more than s_i monsters during each day), or if all monsters are defeated — otherwise, he fights with the next monster. When the hero leaves the dungeon, the current day ends.

Your goal is to defeat the last monster. What is the minimum number of days that you need to achieve your goal? Each day you have to use exactly one hero; it is possible that some heroes don't fight the monsters at all. Each hero can be used arbitrary number of times.

Input

The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of monsters in the dungeon.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the power of the i-th monster.

The third line contains one integer m (1 ≤ m ≤ 2 ⋅ 10^5) — the number of heroes in your party.

Then m lines follow, each describing a hero. Each line contains two integers p_i and s_i (1 ≤ p_i ≤ 10^9, 1 ≤ s_i ≤ n) — the power and the endurance of the i-th hero.

It is guaranteed that the sum of n + m over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case print one integer — the minimum number of days you have to spend to defeat all of the monsters (or -1 if it is impossible).

Example

Input


2
6
2 3 11 14 1 8
2
3 2
100 1
5
3 5 100 2 3
2
30 5
90 1


Output


5
-1
import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef
{       static PrintWriter out=new PrintWriter(System.out);static FastScanner in = new FastScanner(System.in);static class FastScanner {BufferedReader br;StringTokenizer stok;FastScanner(InputStream is) {br = new BufferedReader(new InputStreamReader(is));}
        String next() throws IOException {while (stok == null || !stok.hasMoreTokens()) {String s = br.readLine();if (s == null) {return null;}
                stok = new StringTokenizer(s);}return stok.nextToken();}
        int ni() throws IOException {return Integer.parseInt(next());}long nl() throws IOException {return Long.parseLong(next());}double nd() throws IOException {return Double.parseDouble(next());}char nc() throws IOException {return (char) (br.read());}String ns() throws IOException {return br.readLine();}
        int[] nia(int n) throws IOException{int a[] = new int[n];for (int i = 0; i < n; i++)a[i] = ni();return a;}long[] nla(int n) throws IOException {long a[] = new long[n];for (int i = 0; i < n; i++)a[i] = nl();return a;}
        double[] nda(int n)throws IOException {double a[] = new double[n];for (int i = 0; i < n; i++)a[i] = nd();return a;}int [][] imat(int n,int m) throws IOException{int mat[][]=new int[n][m];for(int i=0;i<n;i++){for(int j=0;j<m;j++)mat[i][j]=ni();}return mat;}
    }
	static long mod=Long.MAX_VALUE;
	public static void main (String[] args) throws java.lang.Exception
	{   int i,j;
	    HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();
		/*  if(hm.containsKey(z))
		        hm.put(z,hm.get(z)+1);
		    else
		        hm.put(z,1);
		 */       
        ArrayList<Integer> arr=new ArrayList<Integer>();
        HashSet<Integer> set=new HashSet<Integer>();
        PriorityQueue<Integer> pq=new PriorityQueue<Integer>();
        int t=in.ni();
        while(t--!=0)
        {   int n=in.ni();
            int a[]=in.nia(n);int m=in.ni();
            int power[]=new int[m];int e[]=new int[m];
            pair p[]=new pair[m];
            for(i=0;i<m;i++)
            {   int x=in.ni();
                int y=in.ni();
                p[i]=new pair(x,y);
                power[i]=x;e[i]=y;
            }
            int temp[]=new int[n+1];
            for(i=0;i<=n;i++)
                temp[i]=-100;
            Arrays.sort(p);
            //for(pair c:p)
            //    out.println(c.x+" "+c.y);
            int counter=0;
            for(i=1;i<=n;i++)
            {   while(counter<m)
                {   //System.out.println("counter="+counter);
                    int z=p[counter].y;
                    if(z<i)
                        counter++; 
                    else
                        break;
                }
                if(counter<m && p[counter].y>=i)
                    temp[i]=p[counter].x;
            }
            //out.println();
            //for(i=1;i<=n;i++)
             //  out.print(temp[i]+" ");
            //out.println();    
            int days=1,flag=0;int max=a[0];counter=1;
            //out.println("temp[1]="+temp[1]);
            for(i=0;i<n;i++)
            {   max=Math.max(a[i],max);
                if(temp[1]<a[i])
                    {flag=1;break;}
                
                if(max<=temp[counter])
                {   //out.println("i="+i);
                    counter++;}
                else
                {   days++;
                    counter=2;
                    max=a[i];
                }
                //out.println("days used="+(counter-1));
            }
            if(flag==1)
                out.println("-1");
            else
                out.println(days);
            //out.println("________________");
        }
        
        
        
        out.close();
	}
	static class pair implements Comparable<pair>{
		int x, y;
		public pair(int x, int y){this.x = x; this.y = y;}
		@Override
		public int compareTo(pair arg0) 
		{   if(x<arg0.x)    return 1;
		    else if(x==arg0.x)
		    {   if(y<arg0.y)    return 1;
		        else if(y>arg0.y)   return -1;
		        else    return 0;
		    }
		    else    return -1;
		}
		public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof pair)) return false;
        pair key = (pair) o;
        return x == key.x && y == key.y;}
        public int hashCode() {int result = x;result = 31 * result + y;return result;}
	}
	static long gcd(long a,long b)
	{   if(b==0)
	        return a;
	    return gcd(b,a%b);    
	}
	static long exponent(long a,long n)
	{   long ans=1;
	    while(n!=0)
	    {   if(n%2==1)
	            ans=(ans*a)%mod;
	       a=(a*a)%mod;
	       n=n>>1;
	    }
	    return ans;
	}
	static int binarySearch(int a[], int item, int low, int high) 
    {   if (high <= low) 
            return (item > a[low])?  (low + 1): low; 
        int mid = (low + high)/2; 
        if(item == a[mid]) 
            return mid+1; 
        if(item > a[mid]) 
            return binarySearch(a, item, mid+1, high); 
        return binarySearch(a, item, low, mid-1); 
    } 
    static void merge(int arr[], int l, int m, int r) {   
        int n1 = m - l + 1; int n2 = r - m; int L[] = new int [n1]; int R[] = new int [n2]; 
        
        for (int i=0; i<n1; ++i) L[i] = arr[l + i]; for (int j=0; j<n2; ++j) R[j] = arr[m + 1+ j]; int i = 0, j = 0; int k = l; while (i < n1 && j < n2) {   if (L[i] <= R[j]) {   arr[k] = L[i]; i++; } else{   arr[k] = R[j]; j++; } k++; } while (i < n1){   arr[k] = L[i]; i++; k++; } while (j < n2) { arr[k] = R[j];   j++;     k++; } 
    } 
    static void Sort(int arr[], int l, int r) {if (l < r) {   int m = (l+r)/2; Sort(arr, l, m); Sort(arr , m+1, r); merge(arr, l, m, r); } } 
    
    static void sort(int a[])
    {Sort(a,0,a.length-1);}
}