【CodeForces - 831C Jury Marks】暴力

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本文探讨了一个有趣的问题：如何根据部分已知的得分变化情况来推断参赛者初始得分的可能性。通过去除重复项并对数据进行排序等操作，文章提供了一种算法解决方案。

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Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.

Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n ≤ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.

Your task is to determine the number of options for the score the participant could have before the judges rated the participant.

Input

The first line contains two integers k and n (1 ≤ n ≤ k ≤ 2 000) — the number of jury members and the number of scores Polycarp remembers.

The second line contains k integers a1, a2, ..., ak ( - 2 000 ≤ ai ≤ 2 000) — jury's marks in chronological order.

The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 ≤ bj ≤ 4 000 000) — the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.

Output

Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).

Examples

input
4 1
-5 5 0 20
10

output
3

input
2 2
-2000 -2000
3998000 4000000

output
1

Note

The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.

In the second example there is only one correct initial score equaling to 4 002 000.

题意：有一个表演有n个评委进行打分，分数sum[i]可能为正可能为负，但是polycarp忘记了初始分是多少，他只记得k个中间的分数b[i]（不按顺序排列），现在问初始分有多少种存在的可能。

分析：首先先对n个分数进行前缀和的处理，记录前i次的分数总和sum[i]，因为b[i]不同，所以把sum[i]中重复的元素去掉，然后设置一个初始的分数s = b[1] - sum[i]，寻找是否存在k个使得b[j] - s == sum[i]的数成立，成立则ans++。注意数组flag[ ]数组的大小，要以分数为下标。

代码如下：

#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long

using namespace std;
const int MX = 2e7 + 5;
const int mod = 1e9 + 7;
const int INF = 2e9 + 5;

int sum[MX];
int flag[MX];
int b[MX];

int main(){
    int n, k;
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++){
        scanf("%d", &sum[i]);
        sum[i] += sum[i-1];
        flag[sum[i]] = 1;
    }
    for(int i = 1; i <= k; i++){
        scanf("%d", &b[i]);
    }
    sort(sum+1, sum+1+n);
    n = unique(sum+1, sum+1+n) - sum - 1;       //去重
    int ans = 0;
    for(int i = 1; i <= n; i++){
        int cnt = 0;
        int s = b[1] - sum[i];
        for(int j = 1; j <= k; j++){
            if(flag[b[j] - s])  cnt++;
        }
        if(cnt == k)    ans++;
    }
    printf("%d\n", ans);
    return 0;
}