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 title "title"
#define ll long long
#define ull unsigned ll
#define fix(x) fixed<<setprecision(x)
#define pii pair<ll,ll>
#define vint vector<ll>
#define pb push_back
using namespace std;
void Freopen(){
	freopen(title".in","r",stdin);
	freopen(title".out","w",stdout);
}
ll read(){
	ll g=0,f=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch){if(ch=='-')f=-1;ch=getchar();}
	while('0'<=ch&&ch<='9'){g=g*10+ch-'0';ch=getchar();}
	return g*f;
}
const ll mod=1e9+7;
const ll N=3005;
ll n,q,a[N],f[N][N],ans;
ll ksm(ll x,ll y){
	ll re=1;
	while(y){
		if(y&1)re=re*x%mod;
		x=x*x%mod;y>>=1;
	}return re;
}
signed main(){
	n=read(),q=read();
	for(ll i=1;i<=n;i++)a[i]=read();
	for(ll i=1;i<=n;i++)for(ll j=1;j<=n;j++)f[i][j]=(a[i]>a[j]);
	ll inv=ksm(2,mod-2),fac=ksm(2,q);
	for(;q--;){
		ll x=read(),y=read();
		f[x][y]=f[y][x]=(f[x][y]+f[y][x])%mod*inv%mod;
		for(ll i=1;i<=n;i++){
			if(i==x||i==y)continue;
			f[i][x]=f[i][y]=(f[i][x]+f[i][y])%mod*inv%mod;
			f[x][i]=f[y][i]=(f[x][i]+f[y][i])%mod*inv%mod;
		}
	}
	for(ll i=1;i<=n;i++)for(ll j=i+1;j<=n;j++)ans=(ans+f[i][j]*fac%mod)%mod;
	return cout<<ans,signed();
}