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

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 entrancen). 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：告诉有几个点，起始位子，走多少步，要你求终点位子.

感觉第一题的题意好难理解。。。

就一步一步模拟就行了

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
	int n,m,k;
	while(cin>>n>>m>>k)
	{
		k=k%n;
		int ans=0;
		if(k==0)
		printf("%d\n",m);
		else
		{
			if(k<0)
			{
				k=-1*k;
				int j=-1;
				for(int i=0;i<k;i++)
				{
					m+=j;
					if(m==0)
					{
					m=n;
					}
					
				}
				
			}
			else
			{
				int j=1;
				for(int i=0;i<k;i++)
				{
					m+=j;
					if(m==n+1)
					{
					m=1;
					}
					
				}
				
			}
			printf("%d\n",m);
		}
		
	}
 } 

别人的代码 （很精简）：

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
	int n,m,k;
	while(cin>>n>>m>>k)
	{
		int ans=0;
		ans=n*100+m+k-1;
		printf("%d\n",ans%n+1);
	}
 }