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 <cstdio>

const int _ = 3e3 + 5, p = 1e9 + 7;

int n, m, a[_], dp[_][_];

inline int power(int x, int k) {
    int res = 1;
    for (; k; k >>= 1, x = 1ll * x * x % p)
        if (k & 1) res = 1ll * res * x % p;
    return res % p;
}

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) dp[i][j] = a[i] < a[j];
    int inv2 = power(2, p - 2);
    for (int x, y, i = 1; i <= m; ++i) {
        scanf("%d %d", &x, &y);
        dp[x][y] = dp[y][x] = 1ll * inv2 * ((dp[x][y] + dp[y][x]) % p) % p;
        for (int j = 1; j <= n; ++j) {
            if (j == x | j == y) continue ;
            dp[x][j] = dp[y][j] = 1ll * inv2 * ((dp[x][j] + dp[y][j]) % p) % p;
            dp[j][x] = dp[j][y] = 1ll * inv2 * ((dp[j][x] + dp[j][y]) % p) % p;
        }
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j < i; ++j) ans = (ans + dp[i][j]) % p;
    ans = 1ll * ans * power(2, m) % p;
    printf("%d\n", ans);
    return 0;
}