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 = 1000000007, inv2 = 500000004;
int p[3010][3010];
int a[3010], n, m;
int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", a + i);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) p[i][j] = a[i] > a[j];
    for (int i = 1; i <= m; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        if (x > y) swap(x, y);
        for (int i = 1; i <= n; i++) {
            if (i == x || i == y) continue;
            p[x][i] = p[y][i] = 1ll * (p[x][i] + p[y][i]) * inv2 % mod;
            p[i][x] = p[i][y] = 1ll * (p[i][x] + p[i][y]) * inv2 % mod;
        }
        p[x][x] = p[y][y] = 0;
        p[x][y] = p[y][x] = 1ll * (p[x][y] + p[y][x]) * inv2 % mod;
    }
    int ans = 0;
    for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++) ans = (ans + p[i][j]) % mod;
    for (int i = 1; i <= m; i++) ans = ans * 2 % mod;
    return cout << ans << endl, 0;
}