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).

Example
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.

題目大意：給出k個評委的評分，這些分數依次加到選手的原始分數上，在這個過程中，某人記得n（n<=k）個不同的分數（順序不定），問原始分數有多少種可能？
解題思路（參考了大神的程式碼）：首先將評委評分字首和排序去重，得到的序列個數為答案可能的最大值（仔細想一下）。那麼開始列舉每個原始成績（即d=c[1]-b[i]），對每一個c[j]都要能夠從d+某個字首和得到，那麼該原始成績就符合條件，否則不符合。

#include <bits/stdc++.h>

using namespace std;
int num[2010],so[2010];
int main()
{
    int k,n;
    cin>>k>>n;
    for(int i=1;i<=k;i++)
    {
        int x;
        scanf("%d",&x);
        num[i]=num[i-1]+x;
    }
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&so[i]);
    }
    sort(num+1,num+k+1);
    int len=unique(num+1,num+k+1)-num-1;
    int ans=len;
    for(int i=1;i<=len;i++)
    {
        int xx=so[1]-num[i];
        for(int j=1;j<=n;j++)
        {
            if(!binary_search(num+1,num+len+1,so[j]-xx))
            {
                ans--;
                break;
            }
        }
    }
    cout<<ans<<endl;

}