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<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=3003;
const int mod=1000000007;
int n,m;
int a[N];
int dp[N][N];
int fpow(int x,int b)
{
	int res=1;
	while(b)
	{
		if(b&1) res=1ll*res*x%mod;
		x=1ll*x*x%mod;
		b>>=1;
	}
	return res;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			dp[i][j]=(a[i]>a[j]);
	int inv=fpow(2,mod-2);
	for(int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		dp[x][y]=dp[y][x]=1ll*(dp[x][y]+dp[y][x])%mod*inv%mod;
		for(int j=1;j<=n;j++)
		{
			if(j==x||j==y) continue;
			dp[x][j]=dp[y][j]=1ll*(dp[x][j]+dp[y][j])%mod*inv%mod;
			dp[j][x]=dp[j][y]=1ll*(dp[j][x]+dp[j][y])%mod*inv%mod;
		}
	}
	int ans=0,num=fpow(2,m);
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			ans=(ans+1ll*dp[i][j]*num%mod)%mod;
	printf("%d",ans);
	return 0;
}