codeforces 659A 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 bcorresponds 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

题意：1~n个数围成一个圆形，a为起点，b为将要走的步数，b>0则顺时针走b步，b<0逆时针走b步问最后终点是哪个点

#include<stdio.h>
#include<string.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<cstdio>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define II __int64
#define DD double
#define MAX 10010
#define mod 10003
#define INF 0x3f3f3f3f
using namespace std;
int main()
{
	int n,m,a,b,i,j;
	while(scanf("%d%d%d",&n,&a,&b)!=EOF)
	{
	//	printf("%d\n",-5%3);
		int sum=0;
		if(b==0) sum=a;
		else if(b<0)
		{
			b=b%n;
			if(a+b<=0)
			    sum=n+a+b;
			else
			    sum=a+b;
		}
		else if(b>0)
		{
			b=b%n;
			if(a+b<=n)
			    sum=a+b;
			else
			    sum=a+b-n;
		}
		printf("%d\n",sum);
	}
	return 0;
}