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 long long mod=1e9+7,inv2=(mod+1)/2;
long long n,q,a[3333],x[3333],y[3333],pw[3333],ipw[3333],dp[3333][3333],cur,ans; 
int main()
{
	scanf("%lld%lld",&n,&q);
	for (long long i=1;i<=n;i++) scanf("%lld",&a[i]);
	for (long long i=1;i<=q;i++) 
	{
		scanf("%lld%lld",&x[i],&y[i]);
		if (x[i]>y[i]) swap(x[i],y[i]);
	}
	pw[0]=1;
	for (long long i=1;i<=3000;i++)
	{
		pw[i]=pw[i-1]*2%mod;
		ipw[i]=ipw[i-1]*inv2%mod;
	}
	for (long long i=1;i<=n;i++)
	{
		for (long long j=i+1;j<=n;j++)
		{
			dp[i][j]=1;
		}
	}
	for (long long i=q;i>=1;i--)
	{
		cur=(dp[x[i]][y[i]]+dp[y[i]][x[i]])*inv2%mod;
		dp[x[i]][y[i]]=dp[y[i]][x[i]]=cur;
		for (long long j=1;j<=n;j++)
		{
			if (j==y[i] || j==x[i]) continue;
			cur=(dp[x[i]][j]+dp[y[i]][j])*inv2%mod;
			dp[x[i]][j]=dp[y[i]][j]=cur;
			cur=(dp[j][x[i]]+dp[j][y[i]])*inv2%mod;
			dp[j][x[i]]=dp[j][y[i]]=cur;
		}
	}
	for (long long i=1;i<=n;i++)
	{
		for (long long j=1;j<=n;j++)
		{
			if (i==j) continue;
			if (a[i]>a[j]) ans=(ans+dp[i][j]*pw[q])%mod;
		}
	}
	printf("%lld\n",ans);
	return 0;
}