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>
#define maxn 3005
#define mod 1000000007
#define count 500000004ll
using namespace std;

int n,q;
int a[maxn];
long long ans;
int dp[maxn][maxn];

inline int read()
{
	int x=0,f=1;
	char ch=getchar();
	while(ch<'0' || ch>'9')
	{
		if(ch=='-')
			f=-1;
		ch=getchar();
	}
	while(ch>='0' && ch<='9')
	{
		x=(x<<1)+(x<<3)+(ch^48);
		ch=getchar();
	}
	return x*f;
}

int work(int x,int k)
{
	int temp=1;
	for(;k;k>>=1,x=1ll*x*x%mod)
		if(k&1)
			temp=1ll*temp*x%mod;
	return temp;
}

int main()
{
	n=read(), q=read();
	for(int i=1;i<=n;i++)
		a[i]=read();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			dp[i][j]=a[i]<a[j];
	
	for(int i=1;i<=q;i++)
	{
		int x=read(), y=read();
		dp[x][y]=dp[y][x]=count*(dp[x][y]+dp[y][x])%mod;
		for(int j=1;j<=n;j++)
			if(j!=x && j!=y)
			{
				dp[j][x]=dp[j][y]=count*(dp[j][x]+dp[j][y])%mod;
				dp[x][j]=dp[y][j]=count*(dp[x][j]+dp[y][j])%mod;
			}
	}
	
	for(int i=1;i<=n;i++)
		for(int j=1;j<i;j++)
			ans+=dp[i][j];
	
	cout<<ans%mod*work(2,q)%mod<<endl;
	
//	system("pause");
	return 0;
}