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;
int n,m,i,j,dp[3005][3005],a[3005],mod=1e9+7,inv=5e8+4,x,y,ans,f=1;
int ag(int x,int y){return 1ll*(x+y)*inv%mod;}
int main()
{
	scanf("%d%d",&n,&m);
	for (i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
	}
	for (i=1;i<=n;i++) for (j=1;j<=n;j++) if (i!=j) dp[i][j]=(a[i]>a[j]);
	while (m--)
	{
		(f*=2)%=mod;
		scanf("%d%d",&x,&y);
		dp[x][y]=dp[y][x]=ag(dp[x][y],dp[y][x]);
		for (i=1;i<=n;i++) if (i!=x&&i!=y)
		{
			dp[x][i]=dp[y][i]=ag(dp[x][i],dp[y][i]);
			dp[i][y]=dp[i][x]=ag(dp[i][y],dp[i][x]);
		}
	}
	for (i=1;i<=n;i++) for (j=i+1;j<=n;j++) (ans+=dp[i][j])%=mod;
	cout<<(1ll*ans*f)%mod<<endl;
	return 0;
}