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 <cstdio>
#include <cstring>
#include <cstdlib>
#include <cassert>
#include <cctype>
#include <algorithm>
#include <functional>
#include <utility>
using ll = long long;
const ll ha = 1000000007LL;
const ll inv_2 = (ha >> 1) + 1LL;
const int maxn = 3005;
ll f[maxn][maxn];
int A[maxn];

int main() {
  int n, Q; 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[i] < A[j]) f[i][j] = 1;
    }
  }
  for(int p = 0; p < Q; p ++) {
    int x, y; scanf("%d%d", &x, &y);
    f[x][y] = f[y][x] = (f[x][y] + f[y][x]) * inv_2 % ha;
    for(int j = 1; j <= n; j ++) {
      if(j != x && j != y) {
        f[j][y] = f[j][x] = (f[j][y] + f[j][x]) * inv_2 % ha;
        f[y][j] = f[x][j] = (f[x][j] + f[y][j]) * inv_2 % ha;
      }
    }
  }
  ll ans = 0;
  for(int i = 1; i <= n; i ++) {
    for(int j = 1; j < i; j ++) {
      ans = (ans + f[i][j]) % ha;
    }
  }
  for(int p = 0; p < Q; p ++) ans = (ans << 1) % ha;
  printf("%lld\n", ans);
  return 0;
}