Codeforces Round #346 (Div. 2) 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.

题意:给你一个圈,分别是1,2,3...,n,然后给a,b(1 ≤ n ≤ 100,1 ≤ a ≤ n,-100 ≤ b ≤ 100),a表示他现在所在的位置,b<0表示逆时针走,b>0表示顺时针走|b|步,问走完之后人所在的位置。

 

思路:如果|b|大于n表示经过一圈之后再走|b|-n步,所以人实际顺时针或者逆时针走的步数为|b|%n,如果是顺时针,答案便为 a+|b|%n > n? a+|b|%n-n : a+|b|%n,

否则答案为 a-|b|%n < 1 ? a-|b|%n+n : a-|b|%n

综合起来答案便是: (a+b)%n，小于0时要加上n

#include<bits/stdc++.h>
using namespace std;

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