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
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
const int N = 3010,mod = 1e9+7,inv=500000004;
int f[N][N],n,Q,a[N];
void upd(int&a,int b){a=1ll*inv*(a+b)%mod;}
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++)
			f[i][j]=a[i]>a[j];
	int K=1;
	for(int i=1;i<=Q;i++) K=K*2%mod;
	while(Q--){
		int x,y;read(x);read(y);
		int res[N*4],cnt=0;
		memset(res,0,sizeof(res));
		for(int i=1;i<=n;i++){
			if(x==i||y==i) continue;
			res[cnt++]=f[y][i],res[cnt++]=f[i][y];
			res[cnt++]=f[x][i],res[cnt++]=f[i][x];
		}
		int now=0;
		for(int i=1;i<=n;i++){
			if(x==i||y==i) continue;
			upd(f[x][i],res[now++]),upd(f[i][x],res[now++]);
			upd(f[y][i],res[now++]),upd(f[i][y],res[now++]);
		}
		res[0]=f[x][y];res[1]=f[y][x];
		upd(f[x][y],res[1]);upd(f[y][x],res[0]);
	}
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			(ans+=f[i][j])%=mod;
	cout<<1ll*K*ans%mod<<'\n';
	return 0;
}