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;
#define ll long long
const int mod=1e9+7, iv2=(mod+1)/2;

int f[3333][3333], g[3333][3333];

int n, q, a[3333];

int main(){
  cin>>n>>q;
  for (int i=1;i<=n;++i) cin>>a[i];
  for (int i=1;i<=n;++i)
    for (int j=1;j<=n;++j)
      if (i!=j) f[i][j]= a[i]>a[j];
  
  ll pw=1;
  for (int x,y;q--;){
    cin>>x>>y;
    for (int i=1;i<=n;++i)
      if (i!=x&&i!=y){
        // i,x  i,y  x,i  y,i
        g[i][x]=(f[i][x]+f[i][y])*(ll)iv2%mod;
        g[i][y]=(f[i][x]+f[i][y])*(ll)iv2%mod;
        g[x][i]=(f[x][i]+f[y][i])*(ll)iv2%mod;
        g[y][i]=(f[x][i]+f[y][i])*(ll)iv2%mod;
      }
    f[x][y]=f[y][x]=(f[x][y]+f[y][x])*(ll)iv2%mod;
    for (int i=1;i<=n;++i)
      if (i!=x&&i!=y){
        // i,x  i,y  x,i  y,i
        f[i][x]=g[i][x];
        f[i][y]=g[i][y];
        f[x][i]=g[x][i];
        f[y][i]=g[y][i];
      }
    pw=pw*2%mod;
  }
  ll ans=0;
  for (int i=1;i<=n;++i)
    for (int j=i+1;j<=n;++j)
      ans=(ans+f[i][j])%mod;
  
  ans=ans*pw%mod;
  cout<<ans<<endl;
}