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;

const int mod = 1e9 + 7, i2 = (mod+1)>>1;

int dp[3000][3000],a[3000];

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n,q;
	cin>>n>>q;
	for(int i=0;i<n;i++)
		cin>>a[i];
	for(int i=0;i<n;i++)
		for(int j=0;j<n;j++)
			if(a[i]>a[j])
				dp[i][j]=1;
	for(int _=1;_<=q;_++){
		int x,y;
		cin>>x>>y;
		x--,y--;
		if(x>y)	swap(x,y);
		int xy = dp[x][y], yx = dp[y][x];
		for(int i=0;i<n;i++){
			dp[x][i]=dp[y][i]=1LL*(dp[x][i]+dp[y][i])*i2%mod;
			dp[i][x]=dp[i][y]=1LL*(dp[i][x]+dp[i][y])*i2%mod;
		}
		dp[x][y]=dp[y][x]=1LL*(xy+yx)*i2%mod;
	}
	long long int ans = 0;
	for(int i=0;i<n;i++)
		for(int j=i+1;j<n;j++)
			ans+=dp[i][j];
	ans%=mod;
	for(int i=0;i<q;i++)
		ans=ans*2%mod;
	cout<<ans<<endl;
}