[Round#424 Div2] C.Jury Marks

http://codeforces.com/problemset/problem/831/C

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 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.

---

k와 N이 주어지고, 이어서 k개의 A배열과 N개의 B배열의 값들이 주어진다.

심사위원들인 A배열 순서대로 점수를 매겼으나 최초 점수를 알 수 없기 때문에 정확한 점수는 알 수가 없다.

B배열은 중간과정의 점수들을 나타낸 것으로, 순서는 중요하지 않다.

이 때, 가능한 최초점수의 경우의수는 몇개인지 구하는 문제

단, B배열의 값들은 모두 다르다는 것이 힌트

A배열순서대로 값을 더하거나 빼기 때문에, 최초 점수가 x라고 하면, x+a1, x+a1+a2, ... 이런식으로 중간과정의 값들이 구해짐을 알 수 있다.

따라서 a1, a1+a2, a1+a2+a3, ... presum을 구해 정렬하고, B배열의 값을 정렬해서 매칭시켰을 때, 

B배열이 모두 들어갈 자리가 있다면 B배열이 전부 중간과정의 값이 됨을 알 수 있다.

#include <stdio.h>
#include <algorithm>
#include <set>
#define MAXN 2004

using namespace std;
set <int> st;
int n, k, C[MAXN], D[MAXN],tmp;
int main() {
	scanf("%d", &n);
	scanf("%d", &k);
	for (int i = 1; i <= n; i++) {
		scanf("%d", C + i);
		C[i] += C[i - 1];
	}
	sort(C + 1, C + n+1);
	for (int i = 1; i <= k; i++) scanf("%d", D + i);
	sort(D + 1, D + k + 1);
	int ans = 0;
	for (int i = 1; i <= n; i++) {//시작점
		int idx = 2;
		int x = C[i];
		if (st.find(x) != st.end()) continue;
		for (int j = i+1 ; j <= n; j++) {
			int p = D[1] + C[j] - x;
			if (D[idx] == p) idx++;
			if (idx == k+1) break;
		}
		if (idx == k + 1) {
			ans++;
			st.insert(x);
		}
	}
	printf("%d", ans);
}