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 pb push_back
const int N=3050;
const int mod=1e9+7;
int pow(int x, int k){ int ans=1;for(;k;k>>=1,x=(ll)x*x%mod) if(k&1) ans=(ll)ans*x%mod;return ans;}
int inv(int x){ return pow(x,mod-2);}
int x[N],y[N],a[N],p[N],sum[N];
int main()
{
	int n,i,q,j;
	scanf("%i %i",&n,&q);
	vector<int> id;
	for(i=1;i<=n;i++) scanf("%i",&a[i]),id.pb(a[i]);
	sort(id.begin(),id.end());id.resize(unique(id.begin(),id.end())-id.begin());
	for(i=1;i<=n;i++) a[i]=lower_bound(id.begin(),id.end(),a[i])-id.begin()+1;
	for(i=1;i<=q;i++) scanf("%i %i",&x[i],&y[i]);
	int ans=0;
	for(j=id.size();j>=1;j--)
	{
		for(i=1;i<=n;i++) p[i]=a[i]==j;
		for(i=1;i<=q;i++) p[x[i]]=p[y[i]]=(ll)(p[x[i]]+p[y[i]])*inv(2)%mod;
		int hi=0;
		for(i=1;i<=n;i++)
		{
			ans+=(ll)hi*p[i]%mod;ans%=mod;
			ans+=(ll)sum[i]*p[i]%mod*inv(2)%mod;ans%=mod;
			hi+=sum[i];hi%=mod;
			sum[i]+=p[i];sum[i]%=mod;
		}
	}
	for(i=1;i<=q;i++) ans=ans*2%mod;
	printf("%i\n",ans);
	return 0;
}