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 N=3050;
typedef long long ll;
const ll mod=1000000007;
const ll inv=500000004;
int n,q,ai[N];
ll dp[N][N],ans,pl;

inline ll KSM(ll a,ll p){
	ll ret=1;
	while(p){
		if(p&1) ret=ret*a%mod;
		a=a*a%mod;
		p>>=1;
	}
	return ret;
}

int main()
{
	scanf("%d%d",&n,&q);
	pl=KSM(2,q);
	for(int i=1;i<=n;i++)
		scanf("%d",&ai[i]);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if(ai[i]>ai[j])
				dp[i][j]=1;
	while(q--){
		int x,y;
		scanf("%d%d",&x,&y);
		if(x>y) swap(x,y);
		dp[x][y]=dp[y][x]=((dp[x][y]+dp[y][x])%mod)*inv%mod;
		for(int i=1;i<=n;i++){
			if(i==x || i==y) continue;
			dp[i][x]=dp[i][y]=((dp[i][y]+dp[i][x])%mod)*inv%mod;
			dp[x][i]=dp[y][i]=((dp[y][i]+dp[x][i])%mod)*inv%mod;
		}
//		for(int i=1;i<=n;i++)	
//		{
//			for(int j=1;j<=n;j++)
//				cout<<dp[i][j]<<" ";
//			cout<<endl;
//		}
	}
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			ans=(dp[i][j]+ans)%mod;
	ans=ans*pl%mod;
	ans=(ans+mod)%mod;
	printf("%lld\n",ans);
	return 0;
}