Codeforces Round #346 (Div. 2)A. Round House

第一场cf比赛，，，今天才写博客0.0

地址：http://codeforces.com/problemset/problem/659/A

题目：

A. Round House

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to n. Entrance n and entrance 1 are adjacent.

Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance a and he decided that during his walk he will move around the house b entrances in the direction of increasing numbers (in this order entrance n should be followed by entrance 1). The negative value of b corresponds to moving |b| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance n). If b = 0, then Vasya prefers to walk beside his entrance.

Illustration for n = 6, a = 2, b =  - 5.

Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.

Input

The single line of the input contains three space-separated integers n, a and b (1 ≤ n ≤ 100, 1 ≤ a ≤ n,  - 100 ≤ b ≤ 100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively.

Output

Print a single integer k (1 ≤ k ≤ n) — the number of the entrance where Vasya will be at the end of his walk.

Examples

input

6 2 -5

output

3

input

5 1 3

output

4

input

3 2 7

output

3

Note

The first example is illustrated by the picture in the statements.

思路：其实就是取模运算，简单题。

ac代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <vector>

using namespace std;

int main (void)
{
        int n,a,b;
        cin>>n>>a>>b;
        while(b < 0)
                b += n;
        b = (b-1)%n+1;
        if(a+b>n)
                a=a+b-n;
        else
                a=a+b;
        cout<<a<<endl;
        return 0;
}