【Codeforces】831C Jury Marks

C. Jury Marks

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 k judges 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).

Examples

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.

题目大意：有一个基础分x，每次对这个基础分加上ki（按顺序），已知在加的过程中出现的中间分（无顺序），问基础分有几种可能

思路：开始想假设每一个已知分的顺序，减去k，仍如map里，但是t了。用前缀和，排序，枚举最小已知分的顺序，若满足条件，放入set里

#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <set>#define maxn 2003using namespace std;int main(){int k, n;int a[maxn] = { 0 }, b[maxn] = { 0 };while (~scanf("%d%d", &k, &n)) {set<int>s;for (int i = 1; i <= k; i++) {            scanf("%d", &a[i]);a[i] += a[i - 1];}for (int i = 1; i <= n; i++)scanf("%d", &b[i]);sort(a + 1, a + 1 + k);sort(b + 1, b + 1 + n);for (int i = 1; i <= k; i++) {int x = b[1] - a[i];int cot = 2;for (int j = i + 1; j <= k; j++) {if (x + a[j] == b[cot]) cot++;}if (cot == n + 1) s.insert(x);}printf("%d\n", s.size());}return 0;}