Vova is playing a computer game. There are in total n turns in the game and Vova really wants to play all of them. The initial charge of his laptop battery (i.e. the charge before the start of the game) is k.

During each turn Vova can choose what to do: 

  * If the current charge of his laptop battery is strictly greater than a, Vova can just play, and then the charge of his laptop battery will decrease by a; 
  * if the current charge of his laptop battery is strictly greater than b (b<a), Vova can play and charge his laptop, and then the charge of his laptop battery will decrease by b; 
  * if the current charge of his laptop battery is less than or equal to a and b at the same time then Vova cannot do anything and loses the game. 



Regardless of Vova's turns the charge of the laptop battery is always decreases.

Vova wants to complete the game (Vova can complete the game if after each of n turns the charge of the laptop battery is strictly greater than 0). Vova has to play exactly n turns. Among all possible ways to complete the game, Vova wants to choose the one where the number of turns when he just plays (first type turn) is the maximum possible. It is possible that Vova cannot complete the game at all.

Your task is to find out the maximum possible number of turns Vova can just play (make the first type turn) or report that Vova cannot complete the game.

You have to answer q independent queries.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 10^5) — the number of queries. Each query is presented by a single line.

The only line of the query contains four integers k, n, a and b (1 ≤ k, n ≤ 10^9, 1 ≤ b < a ≤ 10^9) — the initial charge of Vova's laptop battery, the number of turns in the game and values a and b, correspondingly.

Output

For each query print one integer: -1 if Vova cannot complete the game or the maximum number of turns Vova can just play (make the first type turn) otherwise.

Example

Input


6
15 5 3 2
15 5 4 3
15 5 2 1
15 5 5 1
16 7 5 2
20 5 7 3


Output


4
-1
5
2
0
1

Note

In the first example query Vova can just play 4 turns and spend 12 units of charge and then one turn play and charge and spend 2 more units. So the remaining charge of the battery will be 1.

In the second example query Vova cannot complete the game because even if he will play and charge the battery during each turn then the charge of the laptop battery will be 0 after the last turn.
import java.util.*;
import java.lang.*;
import java.math.*;
import java.awt.Checkbox;
import java.awt.image.ConvolveOp;
import java.io.*;
import java.text.DecimalFormat;
import java.lang.reflect.Array;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigDecimal;
import java.util.*;
public class Codeforces{
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	static long MOD = (long)(1e9+7);
	static FastReader sc = new FastReader();
	static int pInf = Integer.MAX_VALUE;
	static int nInf = Integer.MIN_VALUE;
	public static void main(String[] args){
		int test = 1;
		test = sc.nextInt();
		while(test-->0){
			long n = sc.nextLong();
			long k = sc.nextLong();
			long a = sc.nextLong();
			long b = sc.nextLong();
			long ans = -1;
			long s = 0;
			long e = k;
			while(s<=e) {
				long m = s+(e-s)/2;
				if(check(n,k,a,b,m)) {
					ans = m;
					s = m+1;
				}
				else {
					e = m-1;
				}
			}
			out.println(ans);
  		}
		out.close();
	}
	public static boolean check(long n,long k,long a,long b,long m) {
		return n>(a*m)+((k-m)*b);
	}
	public static long mul(long a, long b){
	    return ((a%MOD)*(b%MOD))%MOD;
	}
	public static long add(long a, long b){
	    return ((a%MOD)+(b%MOD))%MOD;
	}
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//Integer.lowestOneBit(i)   Equals k where k is the position of the first one in the binary
//Integer.highestOneBit(i)  Equals k where k is the position of the last one in the binary
//Integer.bitCount(i)       returns the number of one-bits
//Collections.sort(A,(p1,p2)->(int)(p2.x-p1.x)) To sort ArrayList in descending order wrt values of x.
//            Arrays.parallelSort(a,new Comparator<TPair>() {
//      		public int compare(TPair a,TPair b) {
//        			if(a.y==b.y) return a.x-b.x;
//        			return b.y-a.y;
//        		}
//        	});
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//PrimeFactors    
    public static ArrayList<Long> primeFactors(long n) {
        ArrayList<Long> arr = new ArrayList<>();
        if (n % 2 == 0)
            arr.add((long) 2);
        while (n % 2 == 0)
            n /= 2;
        for (long i = 3; i <= Math.sqrt(n); i += 2) {
            int flag = 0;
            while (n % i == 0) {
                n /= i;
                flag = 1;
            }
            if (flag == 1)
                arr.add(i);
        }
        if (n > 2)
            arr.add(n);
        return arr;
    }
//Pair Class
	static class Pair implements Comparable<Pair>{
		int x;
		int y;
		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
		@Override
		public int compareTo(Pair o) {
			if(this.x==o.x){
			    return (this.y-o.y);
			}
			return (this.x-o.x);
		}
	}
	static class TPair{
		long x;
		int y;
		public TPair(long x, int y) {
			this.x = x;
			this.y = y;
		}
	}
//Merge Sort
	static void merge(long arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        long L[] = new long [n1];
        long R[] = new long [n2];
 
        /*Copy data to temp arrays*/
        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];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        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++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
    static void sort(long arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
//Brian Kernighan’s Algorithm
    static long countSetBits(long n){
        if(n==0) return 0;
        return 1+countSetBits(n&(n-1));
    }
//Factorial
    static long factorial(long n){
        if(n==1) return 1;
        if(n==2) return 2;
        if(n==3) return 6;
        return n*factorial(n-1);
    }
//Euclidean Algorithm
    static long gcd(long A,long B){
        if(B==0) return A;
        return gcd(B,A%B);
    }
//Modular Exponentiation
    static long fastExpo(long x,long n){
        if(n==0) return 1;
        if((n&1)==0) return fastExpo((x*x)%MOD,n/2)%MOD;
        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2))%MOD;
    }
//AKS Algorithm
    static boolean isPrime(long n){
        if(n<=1) return false;
        if(n<=3) return true;
        if(n%2==0 || n%3==0) return false;
        for(int i=5;i<=Math.sqrt(n);i+=6)
            if(n%i==0 || n%(i+2)==0) return false;
        return true;
    }
//Reverse an array
    static <T> void reverse(T arr[],int l,int r){
    	Collections.reverse(Arrays.asList(arr).subList(l, r));
    }
//Sieve of eratosthenes
    static int[] findPrimes(int n){
        boolean isPrime[]=new boolean[n+1];
        ArrayList<Integer> a=new ArrayList<>();
        int result[];
        Arrays.fill(isPrime,true);
        isPrime[0]=false;
        isPrime[1]=false;
        for(int i=2;i*i<=n;++i){
            if(isPrime[i]==true){
                for(int j=i*i;j<=n;j+=i) isPrime[j]=false;
            }
        }
        for(int i=0;i<=n;i++) if(isPrime[i]==true) a.add(i);
        result=new int[a.size()];
        for(int i=0;i<a.size();i++) result[i]=a.get(i);
        return result;
        
    }
//Segmented Sieve
    static boolean[] segmentedSieve(long l, long r){
    	boolean[] segSieve = new boolean[(int)(r-l+1)];
    	Arrays.fill(segSieve, true);
    	int[] prePrimes = findPrimes((int)Math.sqrt(r));
    	for(int p:prePrimes) {
    		long low = (l/p)*p;
    		if(low < l) {
    			low += p;
    		}
    		if(low == p) {
    			low += p;
    		}
    		for(long j = low; j<= r; j += p) {
    			segSieve[(int) (j-l)] = false;
    		}
    	}
    	if(l==1) {
    		segSieve[0] = false;
    	}
    	return segSieve;
    }
//Euler Totent function
    static long countCoprimes(long n){
        ArrayList<Long> prime_factors=new ArrayList<>();
        long x=n,flag=0;
        while(x%2==0){
            if(flag==0) prime_factors.add(2L);
            flag=1;
            x/=2;
        }
        for(long i=3;i*i<=x;i+=2){
            flag=0;
            while(x%i==0){
                if(flag==0) prime_factors.add(i);
                flag=1;
                x/=i;
            }
        }
        if(x>2) prime_factors.add(x);
        double ans=(double)n;
        for(Long p:prime_factors){
            ans*=(1.0-(Double)1.0/p);
        }
        return (long)ans;
    }
	static long modulo = (long)1e9+7;
	public static long modinv(long x){
	    return modpow(x, modulo-2);
	}
	public static long modpow(long a, long b){
	    if(b==0){
	        return 1;
	    }
	    long x = modpow(a, b/2);
	    x = (x*x)%modulo;
	    if(b%2==1){
	        return (x*a)%modulo;
	    }
	    return x;
	}
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception r) {
					r.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception r) {
				r.printStackTrace();
			}
			return str;
		}
	}
}