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;
   #define int long long
   #define PP pair<int,pair<int,int>>
const int MAX = 510000;
const int MOD =  1000000007;
const int INV2 = (MOD+1)/2;
int modpow(int a, int n, int mod) {
    int res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
     signed main(){
   int n,q;
   cin>>n>>q;
   int a[n];
   for(int i=0;i<n;i++)cin>>a[i];
   int dp[n][n],x0=modpow(2,q,MOD);
   for(int i=0;i<n;i++)fill(dp[i],dp[i]+n,0);
   for(int i=0;i<n;i++){
     for(int j=0;j<n;j++){
       if(a[i]>a[j])dp[i][j]=(dp[i][j]+x0)%MOD;
     }
   }
   int x,y;
   for(int p=0;p<q;p++){
     cin>>x>>y;x--;y--;
     int tmp=(dp[y][x]+dp[x][y])%MOD*INV2%MOD;
      dp[y][x]=tmp;dp[x][y]=tmp;
     for(int i=0;i<n;i++)if(i!=x&&i!=y){
      tmp=(dp[i][x]+dp[i][y])%MOD*INV2%MOD;
       dp[i][x]=tmp;dp[i][y]=tmp;
       tmp=(dp[x][i]+dp[y][i])%MOD*INV2%MOD;
       dp[x][i]=tmp;dp[y][i]=tmp;
     }
   }
   int ans=0;
   for(int i=0;i<n;i++)for(int j=i+1;j<n;j++)ans=(ans+dp[i][j])%MOD;
   cout<<ans<<endl;
   
                  
}