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>
#define pb push_back
#define mk make_pair
#define ll long long
#define int long long

using namespace std;

const int maxn = 3010;
const int mod = 1e9+7;

int qsm(int i,int j)
{
	int ans=1;
	while (j)
	{
		if (j&1) ans=ans*i%mod;
		j>>=1;
	    i=i*i%mod;
	}
	return ans;
}

int dp[maxn][maxn];
int n,m;
int q;
int a[maxn];

signed main()
{
   scanf("%lld%lld",&n,&m);
   for (int i=1;i<=n;i++) scanf("%lld",&a[i]);//a[i]=read();
   for (int i=1;i<=n;i++)
     for (int j=1;j<=n;j++)
       dp[i][j]=(a[i]>a[j]);// a[i]在前，a[j]在后形成逆序对的概率 
   int inv = qsm(2,mod-2);
   for (int i=1;i<=m;i++)
   {
   	  int x,y;
   	  scanf("%lld%lld",&x,&y);
   	  dp[x][y]=dp[y][x]=(dp[x][y]+dp[y][x])*inv%mod;
   	  for (int i=1;i<=n;i++)
   	  {
   	  	if (i!=x && i!=y)
   	  	{
   	  		dp[x][i]=dp[y][i]=(dp[x][i]+dp[y][i])*inv%mod;
   	  		dp[i][x]=dp[i][y]=(dp[i][x]+dp[i][y])*inv%mod;
		}
	  }
   }
   int ans=0;
   for (int i=1;i<=n;i++)
     for (int j=i+1;j<=n;j++)
        ans=(ans+dp[i][j])%mod;
   cout<<ans*qsm(2,m)%mod;
   return 0;
}