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 p = 1e9 + 7, inv2 = (p + 1) >> 1;
int n, a[N], m, px[N], py[N];
int f[N][N];
inline void A(int &x, int y) {
    x = (x + y >= p ? x + y - p : x + y);
}
int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", a + i);
    for (int i = 1; i <= m; ++i) scanf("%d%d", px + i, py + i);
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j)
            f[i][j] = i < j;
    for (int i = m; i; --i) {
        int x = px[i], y = py[i];
        for (int j = 1; j <= n; ++j) if (j != y && j != x) {
            f[x][j] = f[y][j] = 1ll * (f[x][j] + f[y][j]) * inv2 % p;
            f[j][y] = f[j][x] = 1ll * (f[j][x] + f[j][y]) * inv2 % p;
        }
        f[x][y] = f[y][x] = 1ll * (f[x][y] + f[y][x]) * inv2 % p;
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j)
            if (a[i] > a[j]) 
                A(ans, f[i][j]);
    while (m--) A(ans, ans);
    printf("%d\n", ans);
    return 0;
}