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 int mod=1000000007,iv2=(mod+1>>1);
int f[3010][3010],a[3010],x,y,n,m,g[3010][3010],nw,ans;

int main()
{
	scanf("%d%d",&n,&m),nw=1;
	for (int i=1; i<=n; i++) scanf("%d",&a[i]);
	for (int i=1; i<=n; i++)
		for (int j=1; j<=n; j++) f[i][j]=(a[i]>a[j]);
	while (m--)
	{
		nw=2ll*nw%mod,scanf("%d%d",&x,&y);
		g[x][x]=g[y][y]=0;
		g[y][x]=g[x][y]=1ll*iv2*(f[x][y]+f[y][x])%mod;
		for (int i=1; i<=n; i++) if (i!=x&&i!=y)
			g[i][y]=g[i][x]=1ll*iv2*(f[i][x]+f[i][y])%mod,
			g[x][i]=g[y][i]=1ll*iv2*(f[x][i]+f[y][i])%mod;
		for (int i=1; i<=n; i++) f[i][x]=g[i][x],f[i][y]=g[i][y],f[x][i]=g[x][i],f[y][i]=g[y][i];
	}
	ans=0;
	for (int i=1; i<=n; i++)
		for (int j=i+1; j<=n; j++) ans=(ans+1ll*nw*f[i][j])%mod;
	return printf("%d\n",ans),0;
}