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;
#define ll long long
const ll Mod=1e9+7;
int n,q,a[3010];
ll f[3010][3010],inv2;
void upd(ll &x,ll v)
{
	x=(x+v)%Mod;
}
ll pw(ll x,ll y)
{
	ll res=1;
	while (y)
	{
		if (y&1)
		{
			res=res*x%Mod;
		}
		x=x*x%Mod;
		y>>=1;
	}
	return res;
}
int main()
{
	inv2=(Mod+1)/2;
	scanf("%d%d",&n,&q);
	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++)
		{
			if (a[i]>a[j])
			{
				f[i][j]=1;
			}
		}
	}
	for (int cas=0;cas<q;cas++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		ll tmp;
		tmp=f[x][y]+f[y][x];
		f[x][y]=f[y][x]=tmp*inv2%Mod;
		for (int i=1;i<=n;i++)
		{
			if (i==x || i==y)
			{
				continue;
			}
			tmp=f[x][i]+f[y][i];
			f[x][i]=f[y][i]=tmp*inv2%Mod;
			tmp=f[i][x]+f[i][y];
			f[i][x]=f[i][y]=tmp*inv2%Mod;
		}
	}
	ll ans=0;
	for (int i=1;i<n;i++)
	{
		for (int j=i+1;j<=n;j++)
		{
			upd(ans,f[i][j]);
		}
	}
	cout<<ans*pw(2,q)%Mod<<endl;
	return 0;
}