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>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
const ll half=(mod+1)/2;
int n,q;
int a[3001];
int x[3001],y[3001];
int p[3001][3001];
ll ax[3001];
ll xb[3001];
ll p2q=1;
int main(){
	ios::sync_with_stdio(false);
	cin >> n >> q;
	for(int i=1; i<=n ;i++) cin >> a[i];
	for(int i=1; i<=q ;i++){
		p2q=p2q*2%mod;
		cin >> x[i] >> y[i];
		if(x[i]>y[i]) swap(x[i],y[i]);
	}
	for(int i=1; i<=n ;i++){
		for(int j=1; j<=n ;j++){
			p[i][j]=(i>j);
		}
	}
	for(int i=q; i>=1 ;i--){
		ll orz=(p[x[i]][y[i]]+p[y[i]][x[i]])*half%mod;
		for(int j=1; j<=n ;j++){
			ax[j]=(p[x[i]][j]+p[y[i]][j])*half%mod;
			xb[j]=(p[j][x[i]]+p[j][y[i]])*half%mod;
		}
		for(int j=1; j<=n ;j++){
			p[x[i]][j]=p[y[i]][j]=ax[j];
			p[j][x[i]]=p[j][y[i]]=xb[j];
		}
		p[x[i]][y[i]]=p[y[i]][x[i]]=orz;
	}
	ll ans=0;
	for(int i=1; i<=n ;i++){
		for(int j=i+1; j<=n ;j++){
			if(a[i]==a[j]) continue;
			if(a[i]<a[j]) ans=(ans+p[i][j])%mod;
			else ans=(ans+mod+1-p[i][j])%mod;
		}
	}
	cout << ans*p2q%mod << endl;
}