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>
#define ll long long
#define ni 500000004
#define N 3010
#define mo 1000000007
using namespace std;
template <typename T> void read(T &x){
	x=0;char c=getchar();int fh=1;
	while (!isdigit(c)){if (c=='-')fh=-1;c=getchar();}
	while (isdigit(c))x=x*10+c-'0',c=getchar();
	x*=fh;
}
ll n,a[N],dp[N][N],q,x,y,ansn;
ll po(ll x,ll y){ll z=1;while (y){if (y%2)z=x*z%mo;x=x*x%mo;y/=2;}return z;}
int main(){
	read(n);read(q);
	for (int i=1;i<=n;i++) read(a[i]);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n;j++)dp[i][j]=(a[i]>a[j]);
	for (int i=1;i<=q;i++){
		read(x);read(y);
		dp[x][y]=dp[y][x]=(dp[x][y]+dp[y][x])*ni%mo;
		for (int j=1;j<=n;j++)
			if (j!=x&&j!=y)dp[j][x]=dp[j][y]=(dp[j][x]+dp[j][y])*ni%mo,dp[x][j]=dp[y][j]=(dp[x][j]+dp[y][j])*ni%mo;
	}
	ll z=po(2,q);
	for (int i=1;i<n;i++)
		for (int j=i+1;j<=n;j++)ansn=(ansn+dp[i][j]*z)%mo;
	cout<<ansn<<endl;
	return 0;
}