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>
#define mod 1000000007
#define i2 500000004ll
using namespace std;
int read();
void Add(int &x, int y) { (x += y) >= mod ? x -= mod : x; }

int n, m;
int a[3003], f[3003][3003];
int main() {
    n = read(), m = read();
    for (int i = 1; i <= n; ++i) {
        a[i] = read();
        for (int j = 1; j < i; ++j)
            f[j][i] = a[j] > a[i], f[i][j] = a[i] > a[j];
    }

    for (int i = 1, x, y; i <= m; ++i) {
        x = read(), y = read();
        for (int j = 1; j <= n; ++j) {
            if (j == x || j == y) continue;
            f[j][x] = f[j][y] = i2 * (f[j][x] + f[j][y]) % mod;
            f[x][j] = f[y][j] = i2 * (f[x][j] + f[y][j]) % mod;
        }
        f[x][y] = f[y][x] = i2 * (f[x][y] + f[y][x]) % mod, swap(f, f);
    }

    int res = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j < i; ++j) Add(res, f[j][i]);

    for (int i = 1; i <= m; ++i) Add(res, res);
    printf("%d\n", res);
    return 0;
}
int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (!isdigit(c)) f = (c == '-') ? -1 : f, c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
    return x * f;
}