Round House （CodeForces - 659A）（绕圈圈，走到哪？）

本文介绍了一个有趣的数学问题——圆屋漫步，通过简单的数学运算帮助读者理解如何确定一个人沿着圆形建筑行走后所处的位置。文章提供了完整的代码实现，并解释了其中涉及到的数学原理。

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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

input

5 1 3

output

input

3 2 7

output

Note

The first example is illustrated by the picture in the statements.

题解

题意简单来说就是：小明住在一个圆形建筑里，有n个出口（第n个出口与第1个出口相连），小明的妈妈嫌他打游戏太久，让他出去遛狗，小明决定从第a个出口开始，沿入口序号的递增方向走b（若b < 0 则沿反方向走），问最后从哪个入口回家。

基础数学题，只要（a+b)%n 就是答案

这里会出现两个问题：① b 可以为负数，若 a+b<0 怎么办，还不知道会少几个n。根据题上给的范围(1 ≤ n ≤ 100, 1 ≤ a ≤ n, - 100 ≤ b ≤ 100)，懒办法就可以在 a+b 后再加 100*n 即可。原式变为（a+b+100*n)%n。

②（a+b)%n 结果为 0 时，表示小明走到了第 n 个出口，而不是第 0 个。

# include <cstdio>

using namespace std;

int main()
{
	int n, a, b;
	scanf("%d%d%d",&n,&a,&b);
	int ans = (a+b+n*100)%n;
	if(!ans) printf("%d\n",n);
	else printf("%d\n",ans);
	return 0;
 }