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.util.StringTokenizer;

public class C {
    public static void main(String[] args) throws IOException {
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        InputReader in=new InputReader(System.in);
        int t=in.nextInt();
        while (t-->0){
            int n=in.nextInt();
            int k=in.nextInt();
            if(n>k){
                out.println(k);
                continue;
            }
            int res= (int) Math.floor(k*1.0/(n-1));
            long cur=res*n;
            long num=res*(n-1);
            if(k%(n-1)!=0)
                out.println((k-num)+cur);
            else out.println((k-num)+cur-1);
        }
        out.flush();
    }
    static class InputReader{
        private BufferedReader in;
        private StringTokenizer tokenizer;
        public InputReader(InputStream stream){
            in = new BufferedReader(new InputStreamReader(stream),32768);
            tokenizer = null;
        }
        public String next() throws IOException{
            while(tokenizer==null || !tokenizer.hasMoreTokens()){
                tokenizer = new StringTokenizer(in.readLine());
            }
            return tokenizer.nextToken();
        }
        public Long nextLong() throws IOException {
            String next = next();
            return Long.valueOf(next);
        }
        public int nextInt() throws IOException{
            return Integer.valueOf(next());
        }
    }
}