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 maxn 3030
#define ll long long
using namespace std;

const ll mod=1e9+7;

ll dp[maxn][maxn],times=1,ans;

int a[maxn],n,T,x,y;

void qpow(int k){
	ll base=2;
	while(k){
		if(k&1) times=(times*base)%mod;
		base=base*base%mod;
		k>>=1;
	}
}

int main(){
	scanf("%d%d",&n,&T);
	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];
	qpow(T);
	while(T--){
		scanf("%d%d",&x,&y);
		for(int i=1;i<=n;i++){
			if(i==x||i==y) continue;
			dp[i][x]=dp[i][y]=(dp[i][x]+dp[i][y])*((mod+1)/2)%mod;
			dp[x][i]=dp[y][i]=(dp[x][i]+dp[y][i])*((mod+1)/2)%mod;
		}
		dp[x][y]=dp[y][x]=(dp[x][y]+dp[y][x])*((mod+1)/2)%mod;
	}
	for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++) ans=(ans+dp[i][j])%mod;
	printf("%lld",ans*times%mod);
	return 0;
}