Codeforces Round #346 (Div. 2)Round House

Vasya lives in a round building, whose entrances are numbered sequentially by integers from1 to n. Entrancen and entrance 1 are adjacent.

Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrancea and he decided that during his walk he will move around the houseb entrances in the direction of increasing numbers (in this order entrancen 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 entrance1 is followed by entrance n). Ifb = 0, then Vasya prefers to walk beside his entrance.

Illustration forn = 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.

Sample Input

Input

6 2 -5

Output

3

Input

5 1 3

Output

4

Input

3 2 7

Output

3题意分析：
给定n，a，b分别表示入口的个数，初始位置以及要走过的路口数，b值为负时逆时针走动，否则则顺时针走动。问最后走到的路口
的标号。
直接模拟：
#include<stdio.h>int main(){int n,a,b,k;while(~scanf("%d%d%d",&n,&a,&b)){if(b<0){b=0-b;k=(a-b)%n;if(k>0)printf("%d\n",k);elseprintf("%d\n",k+n);}else{if((a+b)%n!=0)printf("%d\n",(a+b)%n);elseprintf("%d\n",n);}}}