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<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;

template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

const int mod=1e9+7,N=3005;
const ll inv2=(mod+1)/2;
int f[N][N],g[N],h[N],a[N],n,Q,x,y,pw=1,ans;

void add(int &x,int y){
	x=(x+y>=mod?x+y-mod:x+y);
}

int main(){
	read(n),read(Q);
	rep(i,1,n)read(a[i]);
	rep(i,1,n)rep(j,1,n)
		f[i][j]=(a[i]>a[j]);
	rep(k,1,Q){
		read(x),read(y),pw=pw*2%mod;
		rep(i,1,n)if(i!=x&&i!=y){
			g[i]=inv2*(f[x][i]+f[y][i])%mod;
			h[i]=inv2*(f[i][x]+f[i][y])%mod;
		}
		rep(i,1,n)if(i!=x&&i!=y){
			f[x][i]=f[y][i]=g[i];
			f[i][x]=f[i][y]=h[i];
		}
		f[x][y]=f[y][x]=inv2*(f[x][y]+f[y][x])%mod;
	}
	rep(i,1,n)rep(j,i+1,n)
		add(ans,f[i][j]);
	ans=(ll)ans*pw%mod;
	cout<<ans<<endl;
	return 0;
}