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>
#define MOD 1000000007
#define INV 500000004
using namespace std;
long long f[3005][3005],g[3005],h[3005];
int a[3005];
int main()
{
 int n,q;
 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(a[i]>a[j])f[i][j]=1;
 long long ans=0,tmp=1;
 while(q--)
 {
  int x,y;
  cin>>x>>y;
  for(int i=1;i<=n;i++)
  {
   g[i]=(f[x][i]+f[y][i])*INV%MOD;
   h[i]=(f[i][x]+f[i][y])*INV%MOD;
  }
  for(int i=1;i<=n;i++)
  {
   f[x][i]=f[y][i]=g[i];
   f[i][x]=f[i][y]=h[i];
  }
  tmp=tmp*2%MOD;
  f[x][x]=f[y][y]=0;
  f[x][y]=f[y][x]=(f[x][y]+f[y][x])%MOD;
 }
 for(int i=1;i<=n;i++)
  for(int j=i+1;j<=n;j++)
   ans=(ans+f[i][j])%MOD;
 cout<<ans*tmp%MOD<<endl;
 return 0;
}