Codeforces 659A：Round House（模拟，转化）

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本文介绍了一种计算方法，用于确定一个人从特定入口出发，在一个圆形建筑周围散步一定距离后到达的具体位置。通过分析给定的步数和起始位置，可以有效地计算出最终停留的入口编号。

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.

$\text{[math]}$ 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：一个人散步，给了他的出发点和所走步数，问他最终停在哪个小区外面。

解题思路：根据题意得知，假设n=6，当b为负数时，这个人绕着逆时针走，就用这个序列6 5 4 3 2 1 6 5 4 3 2 1 6 5 4 3 2 1...，从中挑选起点，这么一步步走，当b为正数时，这个人绕着顺时针走，就用这个序列1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6...，从中挑选起点，这么一步步走。

代码如下：

#include <stdio.h>
int main()
{
	int n,a,b;
	scanf("%d%d%d",&n,&a,&b);
	if(b==0)
	printf("%d\n",a);
	if(b>0)//顺时针 
	{
		int yu=b%n;//余下的步数 
		if(a+yu<=n)//余下的步数不足以走过第n个小区 
		{
			printf("%d\n",a+yu);
		}
		else
		{
			yu=yu-(n-a);
			printf("%d\n",yu);
		}
	}
	if(b<0))//逆时针 
	{
		b=-b;
		int bu=b%n;//余下的步数 
		if(bu<a)//序列是降序的，在纸上写写就知道了 
		{
			printf("%d\n",a-bu);
		}
		else
		{
			printf("%d\n",n-(bu-a));
		}
	}
	return 0;
}