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 ff first
#define ss second
#define pb push_back
#define mp make_pair
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int,int> pii;

const int maxn=3111;
const int mod=1e9+7;
int qpow(int x,int y)
{
	int ret=1;
	while(y)
	{
		if(y&1)ret=(LL)ret*x%mod;
		x=(LL)x*x%mod;
		y>>=1;
	}
	return ret;
}
inline void add(int&x,int y){x+=y;if(x>=mod)x-=mod;}
int n,q,a[maxn],w[maxn][maxn];

void do_swap(int u,int v)
{
	if(u>v)swap(u,v);
	
	int tmp=(LL)(w[u][v]+w[v][u])*(mod+1)/2%mod;
	w[u][v]=w[v][u]=tmp;
	
	for(int i=1;i<=n;i++)if(i!=u&&i!=v)
	{
		int tmp;
		tmp=(LL)(w[i][u]+w[i][v])*(mod+1)/2%mod;
		w[i][u]=w[i][v]=tmp;
		
		tmp=(LL)(w[u][i]+w[v][i])*(mod+1)/2%mod;
		w[u][i]=w[v][i]=tmp;
	}
}

int main()
{
	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])w[i][j]=1;
	
	for(int i=1,x,y;i<=q;i++)
	{
		scanf("%d%d",&x,&y);
		do_swap(x,y);
	}
	
	int ans=0;
	for(int i=1;i<=n;i++)for(int j=i+1;j<=n;j++)add(ans,w[i][j]);
	ans=(LL)ans*qpow(2,q)%mod;
	printf("%d\n",ans);
	return 0;
}