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.io.*;
import java.util.*;

import javax.swing.plaf.synth.SynthSeparatorUI;
public class Problem3 {
	static Scanner sc=new Scanner(System.in);
	static PrintWriter out=new PrintWriter(System.out);
	public static void main(String[] args) throws IOException, InterruptedException {
	int q=sc.nextInt();
	while(q-->0) {
	
	int k=sc.nextInt();
	int n=sc.nextInt();
	int a=sc.nextInt();
	int b=sc.nextInt();
	int high=n;
	int low=0;
	int ans=-1;
	while(high>=low) {
		int mid=(high+low)>>1;
		if(check(k,mid,a,b,n)) {
			ans=Math.max(ans, mid);
			low=mid+1;
		}
		else {
			high=mid-1;
		}
		
	}
	System.out.println(ans);
	}
	}
	private static boolean check(int k, int mid, int a, int b,int n) {
		long normal=n-mid;
		long cost1=1l*normal*1l*b*1l;
		long cost2=1l*mid*1l*a*1l;
		if(cost1+cost2>=k)
		return false;
		else
			return true;
	}
	static long pow(long a, int n,int mod)
	{
		long res = 1;
		while(n != 0)
		{
			if((n & 1) == 1)
				res = (res*(a%mod))%mod;
			a *= (a%mod);
			a=a%mod;
			n >>= 1;			
		}
		return res;
	}
	static TreeSet<Integer> primes;
	static int[] isComposite;

	static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new TreeSet<Integer>();

		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				if(1l * i * i <= N)
					for (int j = i * i; j <= N; j += i)	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
			}   
	}
	static int INF=(int) 1E9;
	static class Pair{
		int x;
		int y;
		public Pair(int x,int y) {
			this.x=x;
			this.y=y;
		}
		
	}
	public static int[] schuffle(int a[]) {
		for (int i = 0; i < a.length; i++) {
			int x=(int)(Math.random()*a.length);
		int temp=a[x];
		a[x]=a[i];
		a[i]=temp;
		}
		return a;
	}
	static long gcd(long min,long a) {
		if(a==0) {
			return min;
		}
		return gcd(a,min%a);
	}
static int 	factprime(int n,int p) {
	int pow=0;
	for(int i=p ; i<=n ; i=i*p) {
		pow+=n/i;
	}
	return pow;
}
static long power(long x, long y) 
{ 
    long temp; 
    if( y == 0) 
        return 1; 
       
    temp = power(x, y/2); 
    if (y%2 == 0) 
        return ((temp%10)*(temp)%10)%10; 
    else
        return ((x%10)*(temp%10)*(temp%10))%10;
}  
	static ArrayList<Integer> adj[];
static Stack<Integer> st=new Stack<>();
static  boolean visited[];
static void dfs(int i){
	visited[i]=true;
	st.add(i);
	for(int x:adj[i]) {
		if(!visited[x])
			dfs(x);
	}
}


	static  class Scanner {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s) {
		br = new BufferedReader(new InputStreamReader(s));
	}

	public String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	public int[] nextIntArr(int n) throws IOException {
		int[] arr = new int[n];
		for (int i = 0; i < n; i++)
			arr[i] = nextInt();
		return arr;
	}

	public long[] nextLongArr(int n) throws IOException {
		long[] arr = new long[n];
		for (int i = 0; i < n; i++)
			arr[i] = nextLong();
		return arr;
	}

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

	public String nextLine() throws IOException {
		return br.readLine();
	}

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

	public boolean ready() throws IOException {
		return br.ready();
	}
	public void waitForInput() throws InterruptedException {Thread.sleep(2300);}
}



static class SegmentTree {
	int N; // number of elements in my array after the padding
	int [] array, sTree, lazy; 
	SegmentTree(int []inp){
		N = inp.length - 1;
		array = inp;
		sTree =  new int[N<<1];
		lazy =  new int[N<<1];
		build(1, 1, N);
	}
	void build(int node, int b, int e) {
		if(b == e) {
			//node = b + N - 1
			// update(idx , value)
			// sTree[idx + N - 1] += value
			sTree[node] = array[b];
		}
		else {
			int nodeLeft = node << 1;
			int nodeRight = (node << 1)+1;
			int mid = (b + e) >> 1;
			build(nodeLeft, b , mid);
			build(nodeRight, mid + 1, e);
			sTree[node] = sTree[nodeLeft] + sTree[nodeRight];
		}
	}
	void propagate(int node, int b, int e) {
		int leftChild = node << 1;
		int rightChild = node<<1|1;
		lazy[leftChild] += lazy[node];
		lazy[rightChild] += lazy[node];
		int mid = (b + e)>>1;
		if(lazy[node]%2!=0) {
		sTree[leftChild] = (mid - b + 1)-sTree[leftChild];
		sTree[rightChild] =(e - mid)-sTree[rightChild];
		}
		lazy[node] = 0;
	}
	// add value v for element in position idx in my input array
	//O(log(N))
	void update_point(int idx, int v) {
		int node = idx + N - 1;
		sTree[node] += v;
		while(node > 1) {
			node >>= 1;
			sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
		}
	}
	// O(log(N))
	int query(int i, int j) {
		return query(1, 1, N, i, j); 
	}
	int query(int node, int b, int e, int i, int j) {
		
		
		if(i > e || j < b)
			return 0;
		if(b >= i && e <= j)
			return sTree[node];
		int mid = (b + e) >> 1;
		propagate(node, b, e);
        int q1 = query(node << 1, b, mid, i , j);
		int q2 = query(node<<1|1, mid + 1, e, i, j);
		return q1 + q2;
		
		
	}
	void update_range(int i, int j) { //O(log(N))
		update_range(1, 1, N, i, j); 
	}
	void update_range(int node, int b, int e, int i, int j) {
		if(i > e || j < b)
			return;
		if(b >= i && e <= j) {
			sTree[node] = (e - b + 1)-sTree[node];
			lazy[node]+=1;
		}
		else {
			int mid = (b + e) >> 1;
			propagate(node, b, e);
			update_range(node<<1, b , mid, i, j);
			update_range(node<<1|1, mid+1, e, i, j);
			sTree[node] = sTree[node<<1] + sTree[node<<1|1];
		}
	}
}
}

	

