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 FOR(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int N=3030,P=1e9+7,iv2=(P+1)>>1;
int n,Q,x,y,ans,a[N],f[N][N],g[N][N];
int qpw(int x,int y){int z=1;for(;y;y>>=1,x=1ll*x*x%P)if(y&1) z=1ll*z*x%P;return z;}
int main(){
	scanf("%d%d",&n,&Q);
	FOR(i,1,n) scanf("%d",&a[i]);
	FOR(i,1,n)FOR(j,i+1,n) f[i][j]=a[i]>a[j],g[i][j]=a[i]==a[j];
	FOR(p,1,Q){
		scanf("%d%d",&x,&y);
		if(x>y) swap(x,y);
		f[x][y]=1ll*(1-g[x][y]+P)%P*iv2%P;
		FOR(i,1,x-1){
			f[i][x]=f[i][y]=1ll*(f[i][x]+f[i][y])%P*iv2%P;
			g[i][x]=g[i][y]=1ll*(g[i][x]+g[i][y])%P*iv2%P;
		}
		FOR(i,x+1,y-1){
			int u=f[x][i],v=f[i][y],a=g[x][i],b=g[i][y];
			f[x][i]=1ll*(u+((1-v+P)%P-b+P)%P)%P*iv2%P;
			f[i][y]=1ll*(v+((1-u+P)%P-a+P)%P)%P*iv2%P;
			g[x][i]=g[i][y]=1ll*(a+b)%P*iv2%P;
		}
		FOR(i,y+1,n){
			f[x][i]=f[y][i]=1ll*(f[x][i]+f[y][i])%P*iv2%P;
			g[x][i]=g[y][i]=1ll*(g[x][i]+g[y][i])%P*iv2%P;
		}
		//FOR(i,1,n)FOR(j,1,n) cout<<f[i][j]<<char(j==n?'\n':' ');
	}
	FOR(i,1,n)FOR(j,i+1,n) (ans+=f[i][j])%=P;
	ans=1ll*ans*qpw(2,Q)%P;
	cout<<ans<<'\n';
}