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<cstdio>
#include<iostream>
#define N 3005

using namespace std;

inline int In(){
	char c=getchar(); int x=0,ft=1;
	for(;c<'0'||c>'9';c=getchar()) if(c=='-') ft=-1;
	for(;c>='0'&&c<='9';c=getchar()) x=x*10+c-'0';
	return x*ft;
}

const int P=1e9+7,inv2=5e8+4;
int n,q,A[N],f[N][N],ans=0;

inline int fpow(int x,int k){
	int s=1,t=x;
	for(;k;k>>=1,t=1ll*t*t%P) if(k&1) s=1ll*s*t%P;
	return s;
}

int main(){
	n=In(); q=In();
	for(int i=1;i<=n;++i) A[i]=In();
	for(int i=1;i<=n;++i) for(int j=1;j<=n;++j)
	if(i^j) f[i][j]=(A[i]>A[j]);
	for(int i=1,x,y;i<=q;++i){
		x=In(); y=In(); if(x>y) swap(x,y);
		f[x][y]=f[y][x]=1ll*(f[x][y]+f[y][x])%P*inv2%P;
		for(int j=1;j<=n;++j) if(j^x&&j^y)
		f[x][j]=f[y][j]=1ll*(f[x][j]+f[y][j])%P*inv2%P,
		f[j][x]=f[j][y]=1ll*(f[j][x]+f[j][y])%P*inv2%P;
	}
	for(int i=1;i<=n;++i) for(int j=i+1;j<=n;++j)
	ans=(ans+f[i][j])%P; ans=1ll*fpow(2,q)*ans%P;
	printf("%d\n",ans);
	return 0;
}