Greg has a pad. The pad's screen is an n × m rectangle, each cell can be either black or white. We'll consider the pad rows to be numbered with integers from 1 to n from top to bottom. Similarly, the pad's columns are numbered with integers from 1 to m from left to right.

Greg thinks that the pad's screen displays a cave if the following conditions hold:

  * There is a segment [l, r] (1 ≤ l ≤ r ≤ n), such that each of the rows l, l + 1, ..., r has exactly two black cells and all other rows have only white cells. 
  * There is a row number t (l ≤ t ≤ r), such that for all pairs of rows with numbers i and j (l ≤ i ≤ j ≤ t) the set of columns between the black cells in row i (with the columns where is these black cells) is the subset of the set of columns between the black cells in row j (with the columns where is these black cells). Similarly, for all pairs of rows with numbers i and j (t ≤ i ≤ j ≤ r) the set of columns between the black cells in row j (with the columns where is these black cells) is the subset of the set of columns between the black cells in row i (with the columns where is these black cells). 



Greg wondered, how many ways there are to paint a cave on his pad. Two ways can be considered distinct if there is a cell that has distinct colors on the two pictures.

Help Greg.

Input

The first line contains two integers n, m — the pad's screen size (1 ≤ n, m ≤ 2000).

Output

In the single line print the remainder after dividing the answer to the problem by 1000000007 (109 + 7).

Examples

Input

1 1


Output

0


Input

4 4


Output

485


Input

3 5


Output

451
#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long TESTS = 1;
  while (TESTS--) {
    long long n;
    cin >> n;
    long long adj[n + 1][n + 1], dis[n + 1][n + 1];
    for (long long i = 1; i <= n; i++)
      for (long long j = 1; j <= n; j++)
        cin >> adj[i][j], dis[i][j] = adj[i][j];
    vector<long long> v(n + 1), fans;
    for (long long i = 1; i <= n; i++) cin >> v[i], dis[i][i] = 0;
    reverse(v.begin() + 1, v.end());
    for (long long k = 1; k <= n; k++) {
      long long cur = v[k];
      for (long long i = 1; i <= n; i++)
        for (long long j = 1; j <= n; j++)
          dis[i][j] = min(dis[i][j], dis[i][cur] + dis[cur][j]);
      long long ans = 0;
      for (long long i = 1; i <= k; i++)
        for (long long j = 1; j <= k; j++) ans += dis[v[i]][v[j]];
      fans.push_back(ans);
    }
    reverse(fans.begin(), fans.end());
    for (auto i : fans) cout << i << " ";
  }
  return 0;
}
