Problem: You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.

For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.

Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).

Output

For each test case print the k-th positive integer that is not divisible by n.

Example

Input


6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1


Output


10
15
1999999999
113
1000000001
1

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.*;
import java.util.*;


public class Main{
    static long MOD = 998244353 ;
    static long [] fac;
    static int [] num;

    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        // Start writing your solution here. -------------------------------------
        int t = sc.nextInt();
        //int t = 1;
        while(t-- > 0)
        {
            long n = sc.nextLong();
            long k = sc.nextLong();
            long round = (k - 1) / (n - 1);
            long left = (k - 1) % (n - 1) + 1;
            long res = round * n + left;
            out.println(res);
        }
        out.close();
    }
    /*
    public static void dfs(int ind, int now)
    {
        if(ind > n)
        {
            cnt++;
            int sum = 0;
            for(int i = 0; i < q; i++)
            {
                if(num[b[i]] - num[a[i]] == c[i])
                sum += d[i];
            }
            if(sum > max)
                max = sum;
            return;
        }
        for(int j = now; j <= m; j++)
        {
            num[ind] = j;
            dfs(ind + 1, j);
        }
    }*/
    public static long C(int n, int m)
    {
        if(m == 0 || m == n) return 1l;
        if(m > n || m < 0) return 0l;
        long res = fac[n] * quickPOW((fac[m] * fac[n - m]) % MOD, MOD - 2) % MOD;

        return res;
    }
    public static long quickPOW(long n, long m)
    {
        long ans = 1l;
        while(m > 0)
        {
            if(m % 2 == 1)
                ans = (ans * n) % MOD;
            n = (n * n) % MOD;
            m >>= 1;
        }
        return ans;
    }
    public static int gcd(int a, int b)
    {
        if(a % b == 0) return b;
        return gcd(b, a % b);
    }
    public static long solve(long x, long[] res){
        int n = res.length;
        long a = x / n;
        int b = (int)(x % n);
        return res[n - 1] * a + res[b];
    }
    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;
    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    //--------------------------------------------------------
}