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>
#define N 3001
#define p 1000000007
int a[N],b[N],c[N],f[N][N],i,m,n,u,v,x,y;
int main()
{
	scanf("%d%d",&n,&m);
	for(i=x=1;i<=n;i++)scanf("%d",a+i);
	for(u=1;u<=n;u++)for(v=1;v<=n;v++)if(a[u]>a[v])f[u][v]=1;
	while(m--)
	{
		scanf("%d%d",&u,&v),(x<<=1)%=p,y=500000004ll*(f[u][v]+f[v][u])%p;
		for(i=1;i<=n;i++)b[i]=500000004ll*(f[i][u]+f[i][v])%p,c[i]=500000004ll*(f[u][i]+f[v][i])%p;
		for(i=1;i<=n;i++)f[i][u]=i==v?y:i==u?0:b[i],f[i][v]=i==u?y:i==v?0:b[i],f[u][i]=i==v?y:i==u?0:c[i],f[v][i]=i==u?y:i==v?0:c[i];
	}
	for(i=0,u=1;u<=n;u++)for(v=u;v<=n;v++)i=(i+(long long)x*f[u][v])%p;
	return 0*printf("%d\n",i);
}