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;
typedef long long LL;
LL read(){
	LL x=0,f=0;
	char ch=getchar();
	while (!isdigit(ch))
		f|=ch=='-',ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return f?-x:x;
}
const int N=3005,mod=1e9+7;
int n,q;
int a[N];
int x[N],y[N];
int inv2=(mod+1)/2;
int dp[N][N],tmp[N];
int Pow(int x,int y){
	int ans=1;
	for (;y;y>>=1,x=1LL*x*x%mod)
		if (y&1)
			ans=1LL*ans*x%mod;
	return ans;
}
int main(){
	n=read(),q=read();
	for (int i=1;i<=n;i++)
		a[i]=read();
	memset(dp,0,sizeof dp);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n;j++)
			dp[i][j]=a[i]<a[j]?1:0;
	for (int t=1;t<=q;t++){
		int a=read(),b=read();
		if (a==b)
			continue;
		if (a>b)
			swap(a,b);
		for (int i=1;i<=n;i++){
			if (i==a||i==b)
				continue;
			dp[i][a]=dp[i][b]=(LL)inv2*(dp[i][a]+dp[i][b])%mod;
			dp[a][i]=dp[b][i]=(LL)inv2*(dp[a][i]+dp[b][i])%mod;
		}
		dp[a][b]=dp[b][a]=(LL)inv2*(dp[a][b]+dp[b][a])%mod;
	}
	int ans=0;
	for (int i=1;i<=n;i++)
		for (int j=1;j<i;j++)
			ans=(ans+dp[i][j])%mod;
	ans=(LL)ans*Pow(2,q)%mod;
	cout << ans;
	return 0;
}