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
#include<cstdio>

const int mod=1000000007;
const int inv2=(mod+1)/2;
int n,m,a[5010],x,y,i,j;

long long p[5010][5010],t;
int main()
{
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;++i)scanf("%d",a+i);
	for(i=1;i<=n;++i)for(j=1;j<=n;++j)p[i][j]=a[i]>a[j];
	for (int k=0;k<m;k++) {
		scanf("%d%d",&x,&y);
		p[x][y]=p[y][x]=(p[x][y]+p[y][x])*inv2%mod;
		for(i=1;i<=n;++i)if(i!=x&&i!=y) p[i][x]=p[i][y]=(p[i][x]+p[i][y])*inv2%mod,
			p[x][i]=p[y][i]=(p[x][i]+p[y][i])*inv2%mod;
	}
	for(i=1;i<=n;++i)for(j=n;j>i;--j)t=(t+p[i][j])%mod;
	for (i=0;i<m;++i) t=t*2%mod;
	printf("%lld\n",t);
}