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<iostream>
#include<algorithm>

using namespace std;
typedef long long Int;
#define MOD (1000000007)
Int n, q, x, y;
Int a[3240];
Int dp[3240][3240];

Int pw(Int a, Int n){
  if(n == 0)return 1;
  Int res = pw(a, n/2);
  res = res * res % MOD;
  if(n % 2)res *= a;
  return res % MOD;
}

int main(){
  cin >> n >> q;
  for(Int i = 0;i < n;i++)cin >> a[i];
  for(Int i = 0;i < n;i++){
    for(Int j = 0;j < n;j++){
      dp[i][j] = a[i] < a[j];
    }
  }
  for(Int i = 0;i < q;i++){
    cin >> x >> y;
    x--,y--;
    Int a = dp[x][y] + dp[y][x];
    if(a % 2)a += MOD;
    dp[x][y] = a / 2;
    dp[y][x] = a / 2;
    for(Int j = 0;j < n;j++){
      if(j == x || j == y)continue;
      Int a = (dp[x][j] + dp[y][j]) % MOD;
      if(a % 2)a += MOD;
      a /= 2;
      dp[x][j] = dp[y][j] = a;
    }
    
    for(Int j = 0;j < n;j++){
      if(j == x || j == y)continue;
      Int a = (dp[j][x] + dp[j][y]) % MOD;
      if(a % 2)a += MOD;
      a /= 2;
      dp[j][x] = dp[j][y] = a;
    }
  }
  Int res = 0;
  for(Int i = 0;i < n;i++){
    for(Int j = 0;j < i;j++){
      res += dp[i][j];
      res %= MOD;
    }
  }
  cout << res * pw(2, q) % MOD<< endl;
  return 0;
}