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 md = 1e9 + 7;

inline void add(int &x, int y) {
  x += y;
  if (x >= md) {
    x -= md;
  }
}

inline int mul(int x, int y) {
  return (long long) x * y % md;
}

int main() {
#ifdef wxh010910
  freopen("input.txt", "r", stdin);
#endif
  int n, m;
  scanf("%d %d", &n, &m);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  vector<vector<int>> f(n, vector<int> (n));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      f[i][j] = a[i] > a[j];
    }
  }
  int coef = 1;
  while (m--) {
    coef = mul(coef, 2);
    int x, y;
    scanf("%d %d", &x, &y);
    --x;
    --y;
    f[x][y] = f[y][x] = mul(f[x][y] + f[y][x], md + 1 >> 1);
    for (int i = 0; i < n; ++i) {
      if (i != x && i != y) {
        f[i][x] = f[i][y] = mul(f[i][x] + f[i][y], md + 1 >> 1);
        f[x][i] = f[y][i] = mul(f[x][i] + f[y][i], md + 1 >> 1);
      }
    }
  }
  int answer = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      add(answer, f[i][j]);
    }
  }
  printf("%d\n", mul(answer, coef));
  return 0;
}