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 p=1000000007,inv2=(p+1)/2; 
int qpow(int a,int k){
	int ans=1;
	while(k){
		if(k&1) ans=1LL*ans*a%p;
		a=1LL*a*a%p;
		k>>=1;
	}
	return ans;
}

int N,Q;
int A[3005];
int F[3005][3005];

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++)
		F[i][j]=(A[i]>A[j]);
	for(int i=1;i<=Q;i++){
		int x,y;scanf("%d%d",&x,&y);
		F[x][y]=F[y][x]=1LL*(F[x][y]+F[y][x])%p*inv2%p;
		for(int j=1;j<=N;j++)if(j!=x&&j!=y){
			F[j][x]=F[j][y]=1LL*(F[j][x]+F[j][y])%p*inv2%p;
			F[x][j]=F[y][j]=1LL*(F[x][j]+F[y][j])%p*inv2%p;
		}
	}
	int ans=0;
	for(int i=1;i<=N;i++)
	for(int j=i+1;j<=N;j++)
		ans=(ans+F[i][j])%p;
	printf("%d\n",1LL*ans*qpow(2,Q)%p);
}