A. Round House

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.

 先分成正负来讨论，然后对n取模，再讨论。

#include <iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;

int main()
{
    int n,a,b;
    scanf("%d%d%d",&n,&a,&b);
    if(b>=0){
        b %= n;
        a = (a+b)%n;
        if(a == 0) printf("%d
",n);
       else printf("%d
",a);
    }
    else{
        b = abs(b);
        b %= n;
        if(b<a) printf("%d
",a-b);
        else if(b == a) printf("%d
",n);
        else{
            printf("%d
",n-b+a);
        }
    }
    return 0;
}