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 = 3010;
const int MOD = 1e9 + 7;
int n, q;
int F[N][N], G[N], A[N], inv2 = (MOD + 1) / 2, sum = 1, res;
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)
            if (A[i] < A[j]) F[i][j] = 1;
    while (q --)
    {
        int a, b;
        cin >> a >> b; sum = 2ll * sum % MOD;
        for (int i = 1; i <= n; ++ i) if ((i != a) && (i != b))
            F[a][i] = F[b][i] = 1ll * (F[a][i] + F[b][i]) * inv2 % MOD,
            F[i][a] = F[i][b] = 1ll * (F[i][a] + F[i][b]) * inv2 % MOD;
        F[a][b] = F[b][a] = 1ll * (F[a][b] + F[b][a]) * inv2 % MOD;
    }
    for (int i = 1; i <= n; ++ i)
        for (int j = 1; j < i; ++ j)
            res = (res + F[i][j]) % MOD;
    cout << 1ll * sum * res % MOD << endl;
}