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<iostream>
#include<algorithm>
#define int long long
const int mod=1e9+7;
const int N=3e3+2;
using namespace std;
int dp[N][N],ar[N];
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,i,j,k,l,q,m,ans=0,damn=1,prob=(mod+1)/2;
	cin>>n>>q;
	for(i=1;i<=n;i++){
		cin>>ar[i];
	}
	for(i=1;i<=n;i++){
		for(j=1;j<=n;j++){
			if(ar[i]>ar[j]){
				dp[i][j]=1;
			}
		}
	}
	for(i=1;i<=q;i++){
		damn*=2;
		if(damn>=mod){
			damn-=mod;
		}
		cin>>j>>k;
		for(l=1;l<=n;l++){
			if(j!=l&&k!=l){
				m=dp[j][l];
				dp[j][l]=(dp[j][l]+dp[k][l])*prob;
				dp[j][l]%=mod;
				dp[k][l]=(m+dp[k][l])*prob;
				dp[k][l]%=mod;
				m=dp[l][k];
				dp[l][k]=(dp[l][k]+dp[l][j])*prob;
				dp[l][j]=(m+dp[l][j])*prob;
				dp[l][k]%=mod;
				dp[l][j]%=mod;
			}	
		}
		dp[j][k]=dp[k][j]=(dp[j][k]+dp[k][j])*prob;
		dp[j][k]%=mod;
		dp[k][j]%=mod;
	}
	//cout<<dp[1][3]<<endl;
	for(i=1;i<=n;i++){
		for(j=i+1;j<=n;j++){
			//cout<<i<<' '<<j<<' '<<dp[i][j]<<endl;
			ans+=dp[i][j];
			if(ans>=mod){
				ans-=mod;
			}
		}
	}
	cout<<(ans*damn)%mod;
}