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 <cstdio>
typedef long long ll;
inline int rd(){
    int x=0,p=1;
    char a=getchar();
    while((a<48||a>57)&&a!='-')a=getchar();
    if(a=='-')p=-p,a=getchar();
    while(a>47&&a<58)x=(x<<1)+(x<<3)+(a&15),a=getchar();
    return x*p;
}
const int N=3002;
const ll mod=1e9+7;
int n,q;
int a[N];
ll f[N][N],ans;
inline ll fpow(ll b,ll p=mod-2){
	ll ans=1,tmp=b;
	while(p){
		if(p&1)ans=ans*tmp%mod;
		tmp=tmp*tmp%mod;
		p>>=1;
	}
	return ans;
}
int main(){
	n=rd(),q=rd();
	for(int i=1;i<=n;i++)a[i]=rd();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			f[i][j]=a[i]<a[j];
	ll inv=fpow(2);
	for(int i=1,x,y;i<=q;i++){
		x=rd(),y=rd();
		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){
				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;
			}
	}
	for(int i=1;i<=n;i++)
		for(int j=1;j<i;j++)
			ans=(ans+f[i][j])%mod;
	printf("%lld\n",ans*fpow(2,q)%mod);
    return 0;
}