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
#define mod 1000000007LL

ll f[3010][3010];
int num[3010],n,m;

inline int rd()
{
	int x=0;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar());
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

inline ll ksm(ll x,ll y) { ll res=1;for (;y;y>>=1,x=x*x%mod) if (y&1) res=res*x%mod;return res; }

const ll inv2=ksm(2,mod-2);

int main()
{
	n=rd();m=rd();
	for (int i=1;i<=n;i++) num[i]=rd();
	for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) f[i][j]=(num[i]>num[j]);
	ll hhh=ksm(2,m);
	while (m--)
	{
		int x=rd(),y=rd();
		f[x][y]=f[y][x]=(f[x][y]+f[y][x])*inv2%mod;
		for (int i=1;i<=n;i++)
		{
			if (i==x||i==y) continue;
			f[x][i]=f[y][i]=(f[x][i]+f[y][i])*inv2%mod;
			f[i][x]=f[i][y]=(f[i][x]+f[i][y])*inv2%mod;
		}
	}
	ll ans=0;
	for (int i=1;i<n;i++) for (int j=i+1;j<=n;j++) ans=(ans+f[i][j])%mod;
	printf("%lld\n",ans*hhh%mod);
	return 0;
}