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;
const int N = 512;
int n;
int a[N][N];
int64_t l[N][N];
int64_t ans[N];
int x[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) scanf("%d", &a[i][j]);
  for (int i = 0; i < n; ++i) {
    scanf("%d", x + i);
    --x[i];
  }
  for (int i = n - 1; i >= 0; --i) {
    int xi = x[i];
    for (int j = n - 1; j > i; --j) {
      int xj = x[j];
      l[xj][xi] = a[xj][xi];
      l[xi][xj] = a[xi][xj];
      for (int k = n - 1; k > i; --k) {
        int xk = x[k];
        l[xj][xi] = min(l[xj][xi], l[xj][xk] + a[xk][xi]);
        l[xi][xj] = min(l[xi][xj], a[xi][xk] + l[xk][xj]);
      }
    }
    for (int j = n - 1; j > i; --j) {
      int xj = x[j];
      for (int k = n - 1; k > i; --k) {
        int xk = x[k];
        l[xj][xk] = min(l[xj][xk], l[xj][xi] + l[xi][xk]);
        ans[i] += l[xj][xk];
      }
      ans[i] += l[xi][xj];
      ans[i] += l[xj][xi];
    }
  }
  for (int i = 0; i < n; ++i) {
    printf(
        "%I64d"
        " ",
        ans[i]);
  }
  printf("\n");
  return 0;
}
