A - Round House

A - Round House

Submit Status Practice CodeForces 659A

Description

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

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#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<cctype>#define max(a,b)(a>b?a:b)#define min(a,b)(a<b?a:b)#define INF 0x3f3f3f3f#define N 250000using namespace std;int main(){    int n,a,b,k,i;    while(scanf("%d%d%d",&n,&a,&b)!=EOF)    {        if(b>=0)        {             k=(a+b)%n;             if(k==0)                k=n;        }        else        {            k=a;            for(i=1;i<=-b;i++)            {                k--;                if(k==0)                    k=n;            }        }        printf("%d\n",k);    }    return 0;}

大水题   没找到公式 数据小，循环即可过。