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

                                                       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;
 }