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>
const int mu=1000000007;
const int N=3005,inv2=(mu+1)>>1;
int dp[N][N],a[N],x,y,n,m,ans;
long long ksm(int x,int y){
	long long ans=1;
	for (;y;y>>=1,x=x*1ll*x%mu)
		if (y&1) ans=ans*x%mu;
	return ans;
}
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++)
			if (a[i]>a[j]) dp[i][j]=1;
	for (int i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		dp[x][y]=dp[y][x]=1ll*(dp[x][y]+dp[y][x])*inv2%mu;
		for (int j=1;j<=n;j++){
			if (j==x || y==j) continue;
			dp[y][j]=dp[x][j]=(dp[y][j]+dp[x][j])*1ll*inv2%mu;
			dp[j][y]=dp[j][x]=(dp[j][y]+dp[j][x])*1ll*inv2%mu; 
		}
	}	
	for(int i=1;i<=n;i++)
		for (int j=i+1;j<=n;j++)
			ans=(ans+dp[i][j])%mu;
	printf("%d",ans*ksm(2,m)%mu); 
}