CodeForces 471D MUH and Cube Walls（KMP）

D. MUH and Cube Walls

Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got hold of lots of wooden cubes somewhere. They started making cube towers by placing the cubes one on top of the other. They defined multiple towers standing in a line as a wall. A wall can consist of towers of different heights.

Horace was the first to finish making his wall. He called his wall an elephant. The wall consists of w towers. The bears also finished making their wall but they didn't give it a name. Their wall consists of n towers. Horace looked at the bears' tower and wondered: in how many parts of the wall can he "see an elephant"? He can "see an elephant" on a segment of w contiguous towers if the heights of the towers on the segment match as a sequence the heights of the towers in Horace's wall. In order to see as many elephants as possible, Horace can raise and lower his wall. He even can lower the wall below the ground level (see the pictures to the samples for clarification).

Your task is to count the number of segments where Horace can "see an elephant".

Input

The first line contains two integers n and w (1 ≤ n, w ≤ 2·105) — the number of towers in the bears' and the elephant's walls correspondingly. The second line contains n integers ai (1 ≤ ai ≤ 109) — the heights of the towers in the bears' wall. The third line contains w integers bi (1 ≤ bi ≤ 109) — the heights of the towers in the elephant's wall.

Output

Print the number of segments in the bears' wall where Horace can "see an elephant".

Examples

input

13 5
2 4 5 5 4 3 2 2 2 3 3 2 1
3 4 4 3 2

output

2

题目大意：给出两个序列，代表两堵墙的高度，问在第一个序列中出现了几次第二个序列，第二个序列可以通过将所有数加h或减h来改变高度。

解题思路：无论墙的高度如何变化，两个数之间的差值是不会变的。所以我们可以先预处理出序列相邻两个数的差值，然后在进行KMP匹配，匹配次数即为答案。

代码如下：

#include <bits/stdc++.h>
#define MEM(arr,val) memset(arr,val,sizeof(arr))

const int maxn = 200005;
int next[maxn];
int s[maxn],t[maxn];
int n,w;
int kmp()
{
    MEM(next,-1);
    next[1] = 0;
    int ans = 0;
    for(int i = 2,j = 0;i < w;i++){
        while(j > 0 && s[j + 1] != s[i])
            j = next[j];
        if(s[j + 1] == s[i])
            j += 1;
        next[i] = j;
    }

    for(int i = 1,j = 0;i < n;i++){
        while(j > 0 && s[j + 1] != t[i])
            j = next[j];
        if(s[j + 1] == t[i])
            j += 1;
        if(j == w - 1){
            ans++;
            j = next[j];
        }
    }
    return ans;
}

int main()
{
    int last,cur;
    scanf("%d %d",&n,&w);
    scanf("%d",&last);
    for(int i = 1;i < n;i++){
        scanf("%d",&cur);
        t[i] = cur - last;
        last = cur;
    }
    scanf("%d",&last);
    for(int i = 1;i < w;i++){
        scanf("%d",&cur);
        s[i] = cur - last;
        last = cur;
    }
    printf("%d\n",w == 1 ? n : kmp());
    return 0;
}