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:
// luogu-judger-enable-o2
#include<bits/stdc++.h>
#define nn 3008
#define ll long long
#define mod 1000000007
#define lf double
using namespace std;
ll dp[nn][nn];
ll qpow(ll x,ll y)
{
	ll res=1;
	while(y)
	{
		if(y&1) res=res*x%mod;
		x=x*x%mod;y=y/2;
	}
	return res;
}
ll d2;
int org[nn];
ll tmp1[nn],tmp2[nn];
int main()
{
	d2=qpow(2,mod-2);
	ll n,m;scanf("%lld%lld",&n,&m);
	for(int i=1;i<=n;i++) scanf("%d",&org[i]);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if(org[i]>org[j]) dp[i][j]=1;
	for(int ytl=1;ytl<=m;ytl++)
	{
		int x,y;scanf("%d%d",&x,&y);
		for(int i=1;i<=n;i++) tmp1[i]=(dp[x][i]+dp[y][i])*d2%mod;
		for(int i=1;i<=n;i++) tmp2[i]=(dp[i][x]+dp[i][y])*d2%mod;
		ll tmp3=(dp[x][y]+dp[y][x])*d2%mod;
		for(int i=1;i<=n;i++) 
			dp[i][y]=dp[i][x]=tmp2[i];
		for(int i=1;i<=n;i++)
			dp[x][i]=dp[y][i]=tmp1[i];
		dp[x][x]=dp[y][y]=0;
		dp[x][y]=dp[y][x]=tmp3;
	}
	ll ans=0;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			ans+=dp[i][j],ans%=mod;
	cout<<ans*qpow(2,m)%mod;
    return 0;
}