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;

const int MOD = 1e9 + 7, INV = MOD + 1 >> 1, N = 3333;

int f[N][N], a[N], g[N], h[N], Q, n;

int main(void) {
  scanf("%d%d", &n, &Q);
  for(int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for(int i = 1; i <= n; i++) {
    for(int j = 1; j < n; j++) {
      if(a[j] < a[i]) {
	f[i][j] = 1;
      }
      if(a[j] > a[i]) {
	f[j][i] = 1;
      }
    }
  }
  long long coef = 1;
  while(Q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    for(int i = 1; i <= n; i++) {
      g[i] = 1LL * (f[x][i] + f[y][i]) * INV % MOD;
      h[i] = 1LL * (f[i][y] + f[i][x]) * INV % MOD;
    }
    for(int i = 1; i <= n; i++) {
      if(i == x || i == y) {
	continue;
      }
      f[x][i] = f[y][i] = g[i];
      f[i][y] = f[i][x] = h[i];
    }
    f[x][y] = f[y][x] = 1LL * (f[x][y] + f[y][x]) * INV % MOD;
    coef = coef * 2 % MOD;
  }
  int ans = 0;
  for(int i = 1; i < n; i++) {
    for(int j = i + 1; j <= n; j++) {
      ans = (ans + f[i][j]) % MOD;
    }
  }
  
  cout << ans * coef % MOD << endl;
}