C.Jury Marks

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 a_i 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 kjudges rated the participant there were n (n ≤ k) values b₁, b₂, ..., b_n (it is guaranteed that all values b_j 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, ..., a_k ( - 2 000 ≤ a_i ≤ 2 000) — jury's marks in chronological order.

The third line contains n distinct integers b₁, b₂, ..., b_n ( - 4 000 000 ≤ b_j ≤ 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 2010

Output

3

Input

2 2-2000 -20003998000 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：这道题首先是理解题意，评委给出分数是固定的，但是出现的时间不是固定的，就要知道，哪几种初始分数在和某个或者某几个评委的分数相加之后可以得到给出的记得的分数，这就用到前缀和排序判重，但是还要保证每一个记得的分数都是可以用初始分数和评委的评分得出来的

代码：

#include<iostream>#include<cstring>#include<algorithm>using namespace std;int a[2005];int b[2005];int c[2005];int main(){    int n,m;    while(cin>>n>>m)    {        memset(c,0,sizeof(c));        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        for(int i = 1;i<=n;i++)        {            cin>>a[i];            b[i] = b[i-1]+a[i];        }        sort(b+1,b+n+1);        int help = unique(b+1,b+n+1)-b-1;        for(int i = 1;i<=m;i++)            cin>>c[i];            int re = help;        for(int i = 1;i<=help;i++)        {            int help2 = c[1]-b[i];            for(int j = 1;j<=m;j++)            {                if(!binary_search(b+1,b+n+1,c[j]-help2))                {                    re--;                    break;                }            }        }        cout<<re<<endl;    }}