CodeForces 471D MUH and Cube Walls

题目链接：http://codeforces.com/problemset/problem/471/D

MUH and Cube Walls

time limit per test：2 seconds

memory limit per test：256 megabytes

input：standard input

output：standard output

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 ofw towers. The bears also finished making their wall but they didn't give it a name. Their wall consists ofn 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 ofw 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 containsn integers a_i (1 ≤ a_i ≤ 10⁹) — the heights of the towers in the bears' wall. The third line contains w integers b_i (1 ≤ b_i ≤ 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 52 4 5 5 4 3 2 2 2 3 3 2 13 4 4 3 2

Output

2

Note

The picture to the left shows Horace's wall from the sample, the picture to the right shows the bears' wall. The segments where Horace can "see an elephant" are in gray.

思路：其实就是KMP算法的模板题。关键在于要能够想到相邻两个数作差处理。只要算出全部两个相邻数之间的差，那么剩下的就是裸的KMP算法求匹配串的个数问题了。为什么要作差而不是别的呢？请看下面。

假设第一个数组是a，b，c，第二个数组是d，e，f，那么满足题意就必须满足d-a=e-b=f-c，即e-d=b-a且c-b=f-e，所以要作差处理。注意：m=1时要特殊处理一下。

附上AC代码：

#include <bits/stdc++.h>using namespace std;const int maxn = 200005;int t[maxn], p[maxn], f[maxn];int n, m;void get_fail(){f[0] = f[1] = 0;for (int i=1; i<m; ++i){int j = f[i];while (j && p[i]!=p[j])j = f[j];f[i+1] = p[i]==p[j] ? j+1 : 0;}}int kmp(){get_fail();int j=0, cnt=0;for (int i=0; i<n; ++i){while (j && p[j]!=t[i])j = f[j];if (p[j] == t[i])++j;if (j == m)++cnt;}return cnt;}int main(){ios::sync_with_stdio(false);cin.tie(0);cin >> n >> m;for (int i=0; i<n; ++i)cin >> t[i];for (int i=0; i<m; ++i)cin >> p[i];if (m == 1){cout << n << endl;return 0;}for (int i=0; i<n-1; ++i)t[i] = t[i+1]-t[i];for (int i=0; i<m-1; ++i)p[i] = p[i+1]-p[i];--n, --m;cout << kmp() << endl;return 0;}