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 <cstdio>
#define int long long
#define N 3002
using namespace std;
const int mod=1000000007;
int n,q,i,j,a[N],f[N][N],x[N],y[N],p,inv,ans;
int read()
{
	char c=getchar();
	int w=0;
	while(c<'0'||c>'9') c=getchar();
	while(c<='9'&&c>='0'){
		w=w*10+c-'0';
		c=getchar();
	}
	return w;
}
int poww(int a,int b)
{
	int ans=1,base=a;
	while(b){
		if(b&1) ans=ans*base%mod;
		base=base*base%mod;
		b>>=1;
	}
	return ans;
}
signed main()
{
	n=read();q=read();
	p=poww(2,q),inv=poww(2,mod-2);
	for(i=1;i<=n;i++) a[i]=read();
	for(i=1;i<=q;i++) x[i]=read(),y[i]=read();
	for(i=1;i<=n;i++){
		for(j=1;j<=n;j++) f[i][j]=(a[i]>a[j]);
	}
	for(i=1;i<=q;i++){
		f[x[i]][y[i]]=f[y[i]][x[i]]=(f[x[i]][y[i]]+f[y[i]][x[i]])%mod*inv%mod;
		for(j=1;j<=n;j++){
			if(j==x[i]||j==y[i]) continue;
			f[x[i]][j]=f[y[i]][j]=(f[x[i]][j]+f[y[i]][j])%mod*inv%mod;
			f[j][x[i]]=f[j][y[i]]=(f[j][x[i]]+f[j][y[i]])%mod*inv%mod;
		}
	}
	for(i=1;i<=n;i++){
		for(j=i+1;j<=n;j++) ans=(ans+f[i][j])%mod;
	}
	ans=ans*p%mod;
	printf("%lld\n",ans);
	return 0;
}