Codeforces Round #346 (Div. 2) A.Round House-优快云博客

本文介绍了一个简单的算法问题：一个人居住在一个圆形建筑中，从特定入口出发，按指示方向行走指定的入口数量，最终确定其结束位置的入口编号。通过取模运算解决了无论行走多远都能正确返回最终位置的问题。

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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:给你一个圈,分别是1,2,3...,n,然后给a,b(1 ≤ n ≤ 100,1 ≤ a ≤ n,-100 ≤ b ≤ 100),a表示他现在所在的位置,b<0表示逆时针走,b>0表示顺时针走|b|步,问走完之后人所在的位置。

思路:如果|b|大于n表示经过一圈之后再走|b|-n步,所以人实际顺时针或者逆时针走的步数为|b|%n,如果是顺时针,答案便为 a+|b|%n > n? a+|b|%n-n : a+|b|%n,

否则答案为 a-|b|%n < 1 ? a-|b|%n+n : a-|b|%n

综合起来答案便是: (a+b)%n，小于0时要加上n

#include<bits/stdc++.h>
using namespace std;

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