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<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int read()
{
	char c=getchar();
	int res=0;
	while(c<'0'||c>'9') c=getchar();
	while(c>='0'&&c<='9') res=(res<<1)+(res<<3)+(c^48),c=getchar();
	return res;
}
const int mod=1e9+7,inv2=(mod+1)/2;
int n,q;
int a[3005],f[3005][3005],mi=1,ans=0;
int main()
{
	n=read(),q=read();
	for(int i=1;i<=n;i++) a[i]=read();
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++) f[i][j]=a[i]>a[j];
	while(q--)
	{
		mi=mi*2%mod;
		int x=read(),y=read();
		for(int j=1;j<=n;j++)
		{
			if(j==x||j==y) continue;
			f[x][j]=f[y][j]=1ll*inv2*(f[x][j]+f[y][j])%mod;
			f[j][x]=f[j][y]=1ll*inv2*(f[j][x]+f[j][y])%mod;
		}
		f[x][y]=f[y][x]=1ll*inv2*(f[x][y]+f[y][x])%mod;
	}
	for(int i=1;i<=n;i++)
	for(int j=i+1;j<=n;j++) ans=(ans+f[i][j])%mod;
	ans=1ll*ans*mi%mod;
	printf("%d\n",ans);
}