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 ll long long
using namespace std;
const ll p=1e9+7;
ll n,m,ans,hh,a[3010],f[3010][3010],b[3010][2],c[3010][2],ny;
ll ksm(ll x,ll y){
	ll sej=1;
	while(y){
		if(y&1)sej=sej*x%p;
		x=x*x%p;
		y/=2;
	}
	return sej;
}
int main(){
	ll i,j,x,y;
	scanf("%lld%lld",&n,&m);
	for(i=1;i<=n;i++)scanf("%lld",&a[i]);
	for(i=1;i<=n;i++)
	 for(j=1;j<=n;j++)if(a[i]>a[j])f[i][j]=1;
	ny=ksm(2,p-2);hh=1;
	while(m--){
		hh=hh*2%p;
		scanf("%lld%lld",&x,&y);
		if(x>y)swap(x,y);
		for(i=1;i<=n;i++)b[i][0]=f[i][x],b[i][1]=f[x][i],c[i][0]=f[i][y],c[i][1]=f[y][i];
		for(i=1;i<=n;i++)if(i!=x&&i!=y){
			f[x][i]=(b[i][1]*ny+c[i][1]*ny)%p;
			f[i][x]=(b[i][0]*ny+c[i][0]*ny)%p;
			f[y][i]=(c[i][1]*ny+b[i][1]*ny)%p;
			f[i][y]=(c[i][0]*ny+b[i][0]*ny)%p;
		}
		f[x][y]=f[y][x]=(f[x][y]+f[y][x])*ny%p;
	}
	//printf("%lld\n",hh);
	for(i=1;i<n;i++)
	 for(j=i+1;j<=n;j++)ans=(ans+f[i][j]*hh)%p;
	printf("%lld",ans);
}
/*
f[i,j] a[i]>a[j]的概率 
*/