## Save the Prisoner!

import java.util.Scanner;

public class Solution {

public static void main(String[] args) {

Scanner input = new Scanner(System.in);

int t = input.nextInt(); //number of test cases

for(int i = 1; i <= t; ++i) {

int n = input.nextInt();

//number of prisoners

int m = input.nextInt();

//number of sweets

int s = input.nextInt();

//starting id

//minus one because we count from (include) s

//if s = 2, m = 2

//then the prisoner having poisoined sweet is 3 (not 4)

–m;

m %= n;

System.out.println(m+s > n ? ((m+s)%n) : m+s);

}

}

}

A jail has N prisoners, and each prisoner has a unique id number, S, ranging from 1 to N. There are M sweets that must be distributed to the prisoners.

The jailer decides the fairest way to do this is by sitting the prisoners down in a circle (ordered by ascending S), and then, starting with some random S, distribute one candy at a time to each sequentially numbered prisoner until all M candies are distributed. For example, if the jailer picks prisoner S=2, then his distribution order would be (2,3,4,5,…,n−1,n,1,2,3,4,…) until all M sweets are distributed.

But wait—there’s a catch—the very last sweet is poisoned! Can you find and print the ID number of the last prisoner to receive a sweet so he can be warned?

**Input Format**

The first line contains an integer, T, denoting the number of test cases.

The T subsequent lines each contain 3 space-separated integers:

N (the number of prisoners), M (the number of sweets), and S (the prisoner ID), respectively.

**Constraints**

- 1≤T≤100
- 1≤N≤10^9
- 1≤M≤10^9
- 1≤S≤10^9

**Output Format**

For each test case, print the ID number of the prisoner who receives the poisoned sweet on a new line.

**Sample Input**

```
1
5 2 1
```

**Sample Output**

```
2
```

**Explanation**

There are N=5 prisoners and M=2 sweets. Distribution starts at ID number S=1, so prisoner 1 gets the first sweet and prisoner 2 gets the second (last) sweet. Thus, we must warn prisoner 2 about the poison, so we print 2 on a new line.

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