Codeforces Round #346 (Div. 2)Round House

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.

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);
else
printf("%d\n",k+n);

}
else
{
if((a+b)%n!=0)
printf("%d\n",(a+b)%n);
else
printf("%d\n",n);
}
}
}