Codeforces Round #280 (Div. 2)E good

E. Vanya and Field

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vanya decided to walk in the field of size n × n cells. The field containsm apple trees, the i-th apple tree is at the cell with coordinates (x_i, y_i). Vanya moves towards vector(dx, dy). That means that if Vanya is now at the cell(x, y), then in a second he will be at cell. The following condition is satisfied for the vector:, where is the largest integer that divides botha and b. Vanya ends his path when he reaches the square he has already visited.

Vanya wonders, from what square of the field he should start his path to see as many apple trees as possible.

Input

The first line contains integers n, m, dx, dy(1 ≤ n ≤ 10⁶,1 ≤ m ≤ 10⁵,1 ≤ dx, dy ≤ n) — the size of the field, the number of apple trees and the vector of Vanya's movement. Nextm lines contain integers x_i, y_i (0 ≤ x_i, y_i ≤ n - 1) — the coordinates of apples. One cell may contain multiple apple trees.

Output

Print two space-separated numbers — the coordinates of the cell from which you should start your path.If there are several answers you are allowed to print any of them.

Sample test(s)

Input

5 5 2 30 01 21 32 43 1

Output

1 3

Input

2 3 1 10 00 11 1

Output

0 0

Note

In the first sample Vanya's path will look like: (1, 3) - (3, 1) - (0, 4) - (2, 2) - (4, 0) - (1, 3)

In the second sample: (0, 0) - (1, 1) - (0, 0)

因为路径是循环的，所以可以对y坐标走一遍看每个坐标经过多少步能够回到原来的位置，然后对对应的x坐标进行相应的变化，统计有多少个相同的x，最多的就是答案

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;typedef long long LL;const int maxn=1000010;int N,M,dx,dy;int d[maxn],cnt[maxn];int main(){    while(scanf("%d%d%D%d",&N,&M,&dx,&dy)!=EOF)    {        int y=0,x=0;        for(int i=0;i<N;i++)        {            d[y]=i;            y=(y+dy)%N;        }        memset(cnt,0,sizeof(cnt));        int ans=0,ansx=0;        for(int i=0;i<M;i++)        {            scanf("%d%d",&x,&y);            int tmp=d[y];            tmp=((LL)x+(N-tmp)*(LL)dx)%N;            cnt[tmp]++;            if(cnt[tmp]>ans)            {                ans=cnt[tmp];                ansx=tmp;            }        }        printf("%d %d\n",ansx,0);    }    return 0;}