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;
const int N=3010;
const int mod=1e9+7;
const int inv2=(mod+1)/2;
typedef long long ll;
int n,q;
int a[N];
ll res,dp[N][N];
ll pw(ll x,ll p){
	ll ret=1;
	while(p){
		if(p&1)ret=ret*x%mod;
		x=x*x%mod;
		p>>=1;
	}
	return ret;
}
void pri(){
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++)
			printf("%lld ",dp[i][j]);
		puts("");
	}
}
int main(){
	scanf("%d%d",&n,&q);
	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];
	//pri();
	for(int t=1;t<=q;t++){
		int x,y;scanf("%d%d",&x,&y);
		for(int i=1;i<=n;i++){
			if(i==x||i==y)continue;
			ll t1=dp[x][i],t2=dp[y][i],t3=dp[i][x],t4=dp[i][y];
			dp[x][i]=(t1+t2)%mod*inv2%mod;
			dp[i][x]=(t3+t4)%mod*inv2%mod;
			dp[y][i]=(t1+t2)%mod*inv2%mod;
			dp[i][y]=(t3+t4)%mod*inv2%mod;
		}
		ll t1=dp[x][y],t2=dp[y][x];
		dp[x][y]=(t1+t2)%mod*inv2%mod;
		dp[y][x]=(t1+t2)%mod*inv2%mod;
		//pri();
	}
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			res=(res+dp[i][j])%mod;
	res=res*pw(2,q)%mod;
	res=(res+mod)%mod;
	printf("%lld\n",res);
	return 0;
}
/*
6
1 3 1 2 3 2
*/