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;
#define ri register int
const int N=500005,M=1000000007;
int n,u[N],v[N],i,q,a[N],g[3005][3005],tmp[3005][3005],j;
long long inv,ans;
long long qpow(long long a,long long b)
{
	long long ans=1,s=a;
	while(b)
	{
		if(b&1)
			ans=ans*s%M;
		s=s*s%M;
		b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%d %d",&n,&q);
	for(i=1;i<=n;++i)
		scanf("%d",&a[i]);
	for(i=1;i<=q;++i)
		scanf("%d %d",&u[i],&v[i]);
	for(i=1;i<=n;++i)
		for(j=1;j<=n;++j)
			if(a[i]>a[j])
				g[i][j]=1;
	inv=qpow(2,M-2);
	for(i=1;i<=q;++i)
	{
		for(j=1;j<=n;++j)
			if(j!=u[i]&&j!=v[i])
			{
				tmp[u[i]][j]=tmp[v[i]][j]=(g[v[i]][j]+g[u[i]][j])*inv%M;
				tmp[j][u[i]]=tmp[j][v[i]]=(g[j][u[i]]+g[j][v[i]])*inv%M;
			}
		tmp[v[i]][u[i]]=tmp[u[i]][v[i]]=(g[u[i]][v[i]]+g[v[i]][u[i]])*inv%M;
		for(j=1;j<=n;++j)
		{
			g[u[i]][j]=tmp[u[i]][j];
			g[v[i]][j]=tmp[v[i]][j];
			g[j][u[i]]=tmp[j][u[i]];
			g[j][v[i]]=tmp[j][v[i]];
		}	
	}
	for(i=1;i<=n;++i)
		for(j=i+1;j<=n;++j)
			ans+=g[i][j];
	ans%=M;
	cout<<(ans*qpow(2,q)%M+M)%M;
}