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,q,ans,mod=1e9+7,ress,tmp,a[3333],p[3333],f[3333][3333];
long long quick_q(int x,int y){
	long long res=x,ans=1;
	for(;y;y>>=1,res*=res,res%=mod)if(y&1)ans*=res,ans%=mod;
	return (int)ans;
}
int main(){
	scanf("%d%d",&n,&q),ress=quick_q(2,q),tmp=quick_q(2,mod-2);
	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])f[i][j]=1;
	for(int i=1,x,y;i<=q;f[x][y]=f[y][x]=1ll*(f[x][y]+f[y][x])%mod*tmp%mod,i++){
		scanf("%d%d",&x,&y);
		for(int j=1;j<=n;j++)if(j!=x&&j!=y)f[x][j]=f[y][j]=1ll*(f[x][j]+f[y][j])%mod*tmp%mod;
		for(int j=1;j<=n;j++)if(j!=x&&j!=y)f[j][x]=f[j][y]=1ll*(f[j][x]+f[j][y])%mod*tmp%mod;
	}
	for(int i=1;i<n;i++)for(int j=i+1;j<=n;j++)ans+=1ll*ress*f[i][j]%mod,ans%=mod;
	printf("%d",ans);
}