Codeforces Round #346 (Div. 2) - 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.

題意：1~n的環，從a走b步，會走到那個位置；b>0表示順時針走，b<0表示逆時針走。

分析：水題~~

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define MAXN 100010

int main()
{
    int n,a,b;
    while(cin>>n>>a>>b)
    {
        int t;
        if(b > 0)
            t = (a+b%n)%n;
        else
            t = a-(-b)%n;
        if(t <= 0)t=n+t;
        printf("%d\n",t);
    }
    return 0;
}
</span>