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<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
using namespace std;
#define REP(i,m,n) for(int i=(int)m ; i < (int) n ; ++i )
#define rep(i,n) REP(i,0,n)
typedef long long ll;
typedef pair<int,int> pint;
typedef pair<ll,int> pli;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;

int main(){
    int n,q;
    cin>>n>>q;
    ll p2=1;
    rep(i,q)p2=p2*2%mod;
    ll inv2=(mod+1)/2;
    int a[n];
    rep(i,n)cin>>a[i];
    ll dp[n][n];
    rep(i,n)rep(j,n)dp[i][j]=(a[i]>a[j] ? p2 : 0 );
    rep(i,q){
        int x,y;
        cin>>x>>y;
        --x;--y;
        dp[x][y]=dp[y][x]=(dp[x][y]+dp[y][x])*inv2%mod;
        rep(j,n){
            if(j==x||j==y)continue;
            dp[x][j]=dp[y][j]=(dp[x][j]+dp[y][j])*inv2%mod;
        }
        rep(j,n){
            if(j==x||j==y)continue;
            dp[j][x]=dp[j][y]=(dp[j][x]+dp[j][y])*inv2%mod;
        }
    }
    ll ans=0;
    rep(i,n)rep(j,i)ans+=dp[j][i];
    cout<<ans%mod<<endl;
    return 0;
}