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>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7 , inv2 = (mod+1)/2;
ll n,p=1,q,ans,a[3005],dp[3005][3005];
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    cin>>n>>q;
    for(ll i=1;i<=n;i++)cin>>a[i];
    for(ll i=1;i<=n;i++)for(ll j=1;j<=n;j++)dp[i][j]=a[i]>a[j];
    for(;q>0;q--){
        ll x,y;cin>>x>>y;
        for(ll i=1;i<=n;i++){
            if(i==x || i==y)continue;
            ll p = dp[x][i] + dp[y][i];
            ll q = dp[i][x] + dp[i][y];
            dp[x][i]=dp[y][i]=p*inv2%mod;
            dp[i][x]=dp[i][y]=q*inv2%mod;
        }
        ll r = dp[x][y] + dp[y][x];
        dp[x][y]=dp[y][x]=r*inv2%mod;
        p=(p*2)%mod;
    }
    for(ll i=1;i<=n;i++)for(ll j=i+1;j<=n;j++)ans=(ans + dp[i][j])%mod;
    ans = (ans * p)%mod;
    cout<<ans;
}