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<cassert>
#include<cmath>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<time.h>
#include<algorithm>
using namespace std;
#define REP(i,x,y) for(ll i=x;i<=y;i++)
#define rep(i,n) REP(i,1,n)
#define rep0(i,n) REP(i,0,n-1)
#define repG(i,x) for(ll i=pos[x];~i;i=e[i].next)
#define ll long long
#define db double
const ll N=3007;
const ll mod=1e9+7;
ll Pow(ll x,ll y){
	ll ans=1,now=x;
	while(y){
		if(y&1)ans=ans*now%mod;
		now=now*now%mod;
		y>>=1;
	}
	return ans;
}
ll f[N][N],a[N];
ll n,Q,i2;

int main(){
	scanf("%lld%lld",&n,&Q);
	rep(i,n)scanf("%lld",&a[i]);
	rep(i,n)rep(j,n)if(a[i]>a[j])f[i][j]=1;
	i2=Pow(2,mod-2);
	rep(i,Q){
		ll x,y; scanf("%lld%lld",&x,&y);
		rep(j,n){
			if(j==x||j==y)continue;
			f[j][x]=f[j][y]=(f[j][x]+f[j][y])*i2%mod;
			f[x][j]=f[y][j]=(f[x][j]+f[y][j])*i2%mod;
		}
		f[x][y]=f[y][x]=(f[x][y]+f[y][x])*i2%mod;
	}
	ll ans=0;
	rep(i,n)REP(j,i+1,n)ans=(ans+f[i][j])%mod;
	printf("%lld\n",ans*Pow(2,Q)%mod);
	return 0;
}