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 PYTHON3):

Solution:
#JMD
#Nagendra Jha-4096

 
import sys
import math

#import fractions
#import numpy
 
###File Operations###
fileoperation=0
if(fileoperation):
    orig_stdout = sys.stdout
    orig_stdin = sys.stdin
    inputfile = open('W:/Competitive Programming/input.txt', 'r')
    outputfile = open('W:/Competitive Programming/output.txt', 'w')
    sys.stdin = inputfile
    sys.stdout = outputfile

###Defines...###
mod=1000000007
 
###FUF's...###
def nospace(l):
    ans=''.join(str(i) for i in l)
    return ans
 
 
 
##### Main ####
t=int(input())
for tt in range(t):
    #n=int(input())
    n,k= map(int, sys.stdin.readline().split(' '))
    r = (k//(n-1))
    v = n*r
    d = (k%(n-1))
    ans = v+d
    if d==0:
        ans-=1

    print(ans)
    #a=list(map(int,sys.stdin.readline().split(' ')))
    
    
#####File Operations#####
if(fileoperation):
    sys.stdout = orig_stdout
    sys.stdin = orig_stdin
    inputfile.close()
    outputfile.close()