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 MOD = 1e9 + 7;
const int N = 3e3 + 10;
const int rev2 = (MOD + 1) / 2;

int n, m;
int f[N][N];
int a[N];

int main() {
    cin >> n >> m;
    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]);
    for(int i = 1; i <= m; i++){
        int x, y;
        cin >> x >> y;
        for(int z = 1; z <= n; z++) if (z != x && z != y) {
            f[x][z] = f[y][z] = 1LL * (f[x][z] + f[y][z]) * rev2 % MOD;
            f[z][x] = f[z][y] = 1LL * (f[z][x] + f[z][y]) * rev2 % MOD;
        }
        f[x][y] = f[y][x] = 1LL * (f[x][y] + f[y][x]) * rev2 % MOD;
    }
    long long ans = 0;
    for(int i = 1; i <= n; i++)
        for(int j = i + 1; j <= n; j++) ans = (ans + f[i][j]) % MOD;
    for(int i = 1; i <= m; i++) ans = (ans * 2) % MOD;
    cout << ans;
}