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;
const int N = 3e3 + 11;
const int mod = 1e9 + 7;
int f[N][N], n, q, a[N];
inline int A(int x, int y){
    return x + y - (x + y >= mod ? mod : 0);
}
inline int M(int x, int y){
    return 1LL * x * y % mod;
}
int main(){
    cin>>n>>q;
    for(int i = 1;i <= n; i++)cin>>a[i];
    for(int i = 1;i <= n; i++){
        for(int j = 1;j <= n; j++)f[i][j] = a[i] < a[j];
    }
    int inv2 = (mod + 1) / 2, mi = 1;
    for(int i = 1;i <= q; i++){
        mi = (mi << 1) % mod;
        int x, y;
        cin>>x>>y;
        f[x][y] = f[y][x] = M(inv2, A(f[x][y], f[y][x]));
        for(int j = 1;j <= n; j++){
            if(j == x || j == y)continue;
            f[x][j] = f[y][j] = M(inv2, A(f[x][j], f[y][j]));
            f[j][x] = f[j][y] = M(inv2, A(f[j][x], f[j][y]));
        }
    }
    int ans = 0;
    for(int i = 1;i <= n; i++){
        for(int j = i - 1;j >= 1; j--)ans = A(ans, f[i][j]);
    }
    ans = M(ans, mi);
    cout<<ans<<endl;
    return 0;
}