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<stdio.h>
#include<string.h>
#define P 1000000007
#define N 3200
using namespace std;
long long f[N][N];
long long inv2;
int n,q,a[N];
long long power(long long x,int k){
	long long ans=1;
	while(k){
		if(k&1)(ans*=x)%=P;
		(x*=x)%=P;
		k>>=1;
	}
	return ans;
}
int main(){
	inv2=power(2,P-2);
	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<=i;j++){
			if(a[i]==a[j])continue;
			if(a[i]<a[j])f[i][j]++;
			if(a[j]<a[i])f[j][i]++;
		}
	}
	for(int i=1;i<=q;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		if(x==y)continue;
		long long tmp=(f[x][y]+f[y][x])*inv2%P;
		f[x][y]=tmp;
		f[y][x]=tmp;
		for(int i=1;i<=n;i++){
			if(i==x||i==y)continue;
			long long tmp=(f[x][i]+f[y][i])*inv2%P;
			f[x][i]=tmp;
			f[y][i]=tmp;
			tmp=(f[i][x]+f[i][y])*inv2%P;
			f[i][x]=tmp;
			f[i][y]=tmp;
		}
	}
	long long ans=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<i;j++)(ans+=f[i][j])%=P;
	printf("%lld\n",ans*power(2,q)%P);
}