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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=3005,md=1000000007;
int n,q,i,j,a[N],x[N],y[N],f[N][N],g[N][N],ans,inv2=(md+1)/2;
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",x+i,y+i);
	for(i=1;i<=n;++i)
		for(j=i+1;j<=n;++j)
			f[i][j]=1;
	for(i=q;i>=1;--i)
	{
		for(j=1;j<=n;++j)
		{
			g[x[i]][j]=f[x[i]][j];
			g[y[i]][j]=f[y[i]][j];
			g[j][x[i]]=f[j][x[i]];
			g[j][y[i]]=f[j][y[i]];
		}
		for(j=1;j<=n;++j)
		{
			if(j!=y[i])
			{
				f[x[i]][j]=1ll*(g[x[i]][j]+g[y[i]][j])*inv2%md;
				f[j][x[i]]=1ll*(g[j][x[i]]+g[j][y[i]])*inv2%md;
			}
			if(j!=x[i])
			{
				f[y[i]][j]=1ll*(g[y[i]][j]+g[x[i]][j])*inv2%md;
				f[j][y[i]]=1ll*(g[j][y[i]]+g[j][x[i]])*inv2%md;
			}
		}
		f[x[i]][y[i]]=f[y[i]][x[i]]=1ll*(g[x[i]][y[i]]+g[y[i]][x[i]])*inv2%md;
	}
	for(i=1;i<=n;++i)
		for(j=i+1;j<=n;++j)
			if(a[i]>a[j])
				ans=(ans+f[i][j])%md;
			else if(a[i]<a[j])
				ans=(ans+f[j][i])%md;
	for(i=1;i<=q;++i)
		ans=ans*2ll%md;
	printf("%d",ans);
	return 0;
}