Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

9 5
3
1
4
1
5
9
2
6
5
3 5
8 9
7 9
3 2
3 8


Output

425

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 3002, mod = 1000000007;
int n, q, a[N], f[N][N];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    for (int j = 1; j < i; j++) {
      if (a[j] < a[i]) f[i][j] = 1;
      if (a[j] > a[i]) f[j][i] = 1;
    }
  }
  int ans = 0, t = 1;
  while (q--) {
    (t <<= 1) %= mod;
    int x, y;
    scanf("%d%d", &x, &y);
    f[y][x] = f[x][y] = (ll) (f[x][y] + f[y][x]) * (mod + 1) / 2 % mod;
    for (int i = 1; i <= n; i++) if (i != x && i != y) {
      f[x][i] = f[y][i] = (ll) (f[x][i] + f[y][i]) * (mod + 1) / 2 % mod;
      f[i][x] = f[i][y] = (ll) (f[i][x] + f[i][y]) * (mod + 1) / 2 % mod;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < i; j++) {
      ans = (ans + (ll) f[j][i] * t) % mod;
    }
  }
  printf("%d\n", ans);
}