time limit per test
1 second
memory limit per test
256 megabytes
Blackslex is designing a log-in system for Gean Dev and discovered that most users use weak passwords.
To resolve this issue, he posed the following conditions, dependent on two variables k and x, for all passwords. Each password is a string s of length n satisfying these properties.
- s uses only the first k lowercase letters of the English alphabet.
- For every pair of indices 1≤i<j≤n such that (j−i) is divisible by x, the letters si and sj are different.
Find the smallest integer n such that no valid string of length n exists.
Input
The first line contains a single integer t (1≤t≤500) --- the number of test cases.
The first and only line of each test case contains two integers k and x (1≤k≤26, 1≤x≤15).
Output
For each test case, output the minimal n.
Example
Input
Copy
3
2 1
3 2
1 5
Output
Copy
3
7
6
Note
For the first test case, there are no valid strings of length n=3. For n=2, one such valid example is ab. Note that the only pair (i,j) that (j−i) is divisible by x=1 and 1≤i<j≤n for n=2 is (1,2).
For the second test case, there are no valid strings of length n=7. For n=6, one such valid example is aabccb. Note that all pairs (i,j) that (j−i) is divisible by x=2 and 1≤i<j≤n for n=6 include (1,3), (1,5), (2,4), (2,6), (3,5), and (4,6).
For the third test case, there are no valid strings of length n=6. For n=5, one such valid example is aaaaa. Note that there are no pairs (i,j) that (j−i) is divisible by x=5 and 1≤i<j≤n for n=5.
解题说明：此题是一道数学题，找规律能发现输出k*x+1。求出来满足有效字符串的最大的n，将其+1即可，让每个字符连续出现x次，答案即为k*x+1。
cpp
#include<stdio.h>
int main()
{
int t;
scanf("%d", &t);
for (int i = 0; i < t; i++)
{
int k, x;
scanf("%d %d", &k, &x);
printf("%d\n", 1 + k * x);
}
return 0;
}