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>
//dengyaotriangle
using namespace std;

const int maxn=3005;
const int mdn=1000000007;
const int inv2=(mdn+1)>>1;
int n,q;
int a[maxn];
pair<int,int> c[maxn];
long long dp[maxn][maxn];

int main(){
    ios::sync_with_stdio(0);
    cin>>n>>q;
    for(int i=1;i<=n;i++){cin>>a[i];}
    for(int i=1;i<=q;i++){cin>>c[i].first>>c[i].second;}
    for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){dp[i][j]=(a[i]<a[j]);}} 
    for(int i=1;i<=q;i++){
        int u=c[i].first,v=c[i].second;
        long long mean=(dp[u][v]+dp[v][u])*inv2%mdn;
        dp[u][v]=dp[v][u]=mean;
        for(int j=1;j<=n;j++){
            if(j==u||j==v)continue;
            long long m1=(dp[j][u]+dp[j][v])*inv2%mdn;
            dp[j][v]=m1;
            dp[j][u]=m1;
            long long m2=(dp[u][j]+dp[v][j])*inv2%mdn;
            dp[u][j]=m2;
            dp[v][j]=m2;
        }
    }
    long long ans=0;
    for(int i=1;i<=n;i++){
        for(int j=1;j<i;j++){
            ans+=dp[i][j];
        }
    }
    long long tot=1;
    for(int i=1;i<=q;i++){tot=(tot<<1)%mdn;}
    cout<<(ans%mdn*tot)%mdn;
    return 0;
}