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;
int dx[8] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[8] = {0, 0, -1, 1, 1, -1, -1, 1};
const int N = 505;
int dis[N][N], n, k, bad[N], kk[N];
long long ans[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      scanf("%d", &dis[i][j]);
    }
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &kk[i]);
    bad[kk[i] - 1] = 1;
  }
  for (int i = n - 1; i > -1; i--) {
    k = kk[i] - 1;
    bad[k] = 0;
    for (int u = 0; u < n; u++) {
      for (int v = 0; v < n; v++) {
        dis[u][v] = min(dis[u][v], dis[u][k] + dis[k][v]);
      }
    }
    for (int u = 0; u < n; u++) {
      for (int v = 0; v < n; v++) {
        if (!bad[u] && !bad[v] && u != v) {
          ans[i] += dis[u][v];
        }
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (i) {
      printf(" ");
    }
    printf("%lld", ans[i]);
  }
  puts("");
  return 0;
}
