A. Round House

A. 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.

先分成正负来讨论，然后对n取模，再讨论。

 #include <iostream>
 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 using namespace std;

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

卷珠帘