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<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=3010,mod=1e9+7,inv2=5e8+4;
int n,Q,x,y,a[N];
ll dp[N][N],t1,t2,t3,t4;
ll ans,tot=1;
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++)if (a[i]>a[j]) dp[i][j]=1;
	while (Q--)
	{
		tot=tot*2%mod;
		scanf("%d%d",&x,&y);
		dp[x][y]=dp[y][x]=((ll)dp[x][y]*inv2%mod+(ll)dp[y][x]*inv2%mod)%mod;
		for (int i=1;i<=n;i++) 
		if (i!=x&&i!=y){
			t1=dp[x][i];t2=dp[i][x];t3=dp[y][i];t4=dp[i][y];
			dp[y][i]=dp[x][i]=((ll)t1*inv2%mod+(ll)t3*inv2%mod)%mod;
			dp[i][x]=dp[i][y]=((ll)t2*inv2%mod+(ll)t4*inv2%mod)%mod;
		}
	}
	for (int i=1;i<n;i++) for (int j=i+1;j<=n;j++) ans=(ans+dp[i][j])%mod;
	ans=ans*tot%mod;
	cout<<ans;
}