A - Round House

A - Round House

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Description

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.

Sample Input

Input

6 2 -5

Output

3

Input

5 1 3

Output

4

Input

3 2 7

Output

3

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<cctype>
#define max(a,b)(a>b?a:b)
#define min(a,b)(a<b?a:b)
#define INF 0x3f3f3f3f

#define N 250000
using namespace std;

int main()
{
    int n,a,b,k,i;
    while(scanf("%d%d%d",&n,&a,&b)!=EOF)
    {
        if(b>=0)
        {
             k=(a+b)%n;
             if(k==0)
                k=n;
        }
        else
        {
            k=a;
            for(i=1;i<=-b;i++)
            {
                k--;
                if(k==0)
                    k=n;
            }
        }
        printf("%d\n",k);
    }
    return 0;
}

大水题   没找到公式 数据小，循环即可过。