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 N=3e3+5,mo=1e9+7,iv2=mo+1>>1;
int n,q,a[N],x,y,f[N][N],i,j,ans,mi=1;
int main(){
	scanf("%d%d",&n,&q);
	for(i=1;i<=n;++i)scanf("%d",a+i);
	for(i=1;i<=n;++i)for(j=1;j<=n;++j)f[i][j]=a[i]>a[j];
	for(;q--;){
		static int g1[N],g2[N],g3[N],g4[N];
		scanf("%d%d",&x,&y);
		for(i=1;i<=n;++i)g1[i]=f[i][x],g2[i]=f[x][i],g3[i]=f[i][y],g4[i]=f[y][i];
		j=1ll*iv2*(f[x][y]+f[y][x])%mo;
		for(i=1;i<=n;++i){
			f[i][x]=1ll*iv2*(f[i][x]+g3[i])%mo;
			f[x][i]=1ll*iv2*(f[x][i]+g4[i])%mo;
			f[i][y]=1ll*iv2*(f[i][y]+g1[i])%mo;
			f[y][i]=1ll*iv2*(f[y][i]+g2[i])%mo;
		}
		f[y][x]=f[x][y]=j;mi=mi*2%mo;f[x][x]=f[y][y]=0;
	}
	for(i=1;i<n;++i)for(j=i+1;j<=n;++j)ans=(ans+f[i][j])%mo;
	ans=1ll*ans*mi%mo;
	printf("%d\n",ans);
	return 0;
}