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<algorithm>
#include<cstring>
#include<cctype>
#include<cstdio>
#define rep(i,x,y) for(int i=x; i<=y; ++i)
#define repd(i,x,y) for(int i=x; i>=y; --i)

using namespace std;
const int N=3005,mod=1000000007;
typedef long long LL;
const LL inv2=(mod+1)/2;
int n,m,a[N];
struct data{int x,y;} dat[N];
LL f[N][N],g[N][N],A,ans;

int getint()
{	
	char ch;
	while(!isdigit(ch=getchar()));
	int x=ch-48;
	while(isdigit(ch=getchar())) x=x*10+ch-48;
	return x;
}

void inc(LL &x,LL y)
{
	x=(x+y)*inv2%mod;
}

int main()
{
	n=getint(),m=getint();
	rep(i,1,n) a[i]=getint();
	rep(i,1,m) dat[i].x=getint(),dat[i].y=getint();
	rep(i,1,n) rep(j,1,i-1) f[i][j]=1;
	A=1;
	repd(i,m,1)
	{
		int x=dat[i].x,y=dat[i].y;
		rep(j,1,n)
		{
			g[x][j]=f[x][j];
			g[y][j]=f[y][j];
			g[j][x]=f[j][x];
			g[j][y]=f[j][y];
		}
		rep(j,1,n) if(j!=x && j!=y)
		{
			inc(f[y][j],g[x][j]),inc(f[j][y],g[j][x]);
			inc(f[x][j],g[y][j]),inc(f[j][x],g[j][y]);
		}
		inc(f[x][y],g[y][x]),inc(f[y][x],g[x][y]);
		A=A*2%mod;
	}
	rep(i,1,n) rep(j,1,n) if(a[i]<a[j]) ans=(ans+f[i][j])%mod;
	printf("%lld\n",ans*A%mod);
	return 0;
}