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<cstdlib>
#include<cstring>

using namespace std;

const int maxn=3010;
const int mo=1000000007;

int n,m,f[maxn][maxn],z[maxn];

int mul(int a,int b)
{
	int ans=1;
	while (b)
	{
		if (b&1) ans=1ll*ans*a%mo;
		a=1ll*a*a%mo;
		b>>=1;
	}
	return ans;
}

int main()
{
	scanf("%d%d",&n,&m);
	for (int a=1;a<=n;a++)
		scanf("%d",&z[a]);
	for (int a=1;a<=n;a++)
		for (int b=1;b<=n;b++)
			f[a][b]=(z[a]<z[b]);
	int div2=mul(2,mo-2);
	for (int a=1;a<=m;a++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		if (x==y) continue;
		f[x][y]=f[y][x]=(0ll+f[x][y]+f[y][x])*div2%mo;
		for (int b=1;b<=n;b++)
			if (b!=x && b!=y)
			{
				f[b][x]=f[b][y]=(0ll+f[b][x]+f[b][y])*div2%mo;
				f[x][b]=f[y][b]=(0ll+f[x][b]+f[y][b])*div2%mo;
			}
	}
	int ans=0;
	for (int a=1;a<=n;a++)
		for (int b=a+1;b<=n;b++)
		{
			ans += f[b][a];
			if (ans>=mo) ans-=mo;
		}
	printf("%d\n",1ll*ans*mul(2,m)%mo);

	return 0;
}