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 rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

ll ppow(ll a,ll b){
	ll res=1;
	while(b){
		if(b&1)res=(res*a)%MOD;
		a=(a*a)%MOD;
		b>>=1;
	}
	return res;
}

ll E[4000][4000];
int a[4000];

int main(){
	ll t=ppow(2,MOD-2);
	int n,q;scanf("%d%d",&n,&q);
	rep(i,n)scanf("%d",&a[i]);
	rep(i,n)rep(j,n){
		E[i][j]=(a[i]>a[j]);
	}
	rep(i,q){
		int x,y;scanf("%d%d",&x,&y);x--;y--;
		if(x>y)swap(x,y);
		rep(j,n){
			if(j==x||j==y)continue;
			E[j][x]=E[j][y]=(E[j][x]*t+E[j][y]*t)%MOD;
			E[x][j]=E[y][j]=(E[x][j]*t+E[y][j]*t)%MOD;
		}
		E[x][y]=E[y][x]=(E[x][y]*t+E[y][x]*t)%MOD;
	}
	ll ans=0;
	rep(i,n){
		for(int j=i+1;j<n;j++){
			(ans+=E[i][j])%=MOD;
		}
	}
	cout<<ans*ppow(2,q)%MOD<<endl;
}