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 <cstring>
#include <cstdio>
#define ll long long
using namespace std;

inline int read(){
	int sum=0,f=1;char c=getchar();
	while(c>'9'||c<'0'){if(c=='-') f=-1;c=getchar();}
	while(c<='9'&&c>='0'){sum=sum*10+c-'0';c=getchar();}
	return f*sum;
}

const int N=5005;
const int Mod=1000000007;
int n,m,a[N];
ll f[N][N];

inline ll fpow(ll a,ll b){
	ll ret=1;
	while(b){if(b&1) ret=ret*a%Mod;a=a*a%Mod;b>>=1;}
	return ret;
}

signed main(){
	n=read();m=read();ll inv=fpow(2,Mod-2);
	for(int i=1;i<=n;i++) a[i]=read();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++){
			if(i==j) continue;
			if(a[i]>a[j]) f[i][j]=1;
		}
	for(int i=1;i<=m;i++){		
		int x=read(),y=read();
		f[x][y]=f[y][x]=(f[x][y]+f[y][x])%Mod*inv%Mod;
		for(int j=1;j<=n;j++){
			if(j==x||j==y) continue;			
			f[x][j]=f[y][j]=(f[x][j]+f[y][j])%Mod*inv%Mod;
			f[j][x]=f[j][y]=(f[j][x]+f[j][y])%Mod*inv%Mod;
		}
	}
	ll ans=0;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++) ans=(ans+f[i][j])%Mod;
	printf("%lld\n",ans*fpow(2,m)%Mod);
	return 0;
}