p03189 AtCoder Grand Contest 030 - Inversion Sum
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
```python
import sys

n, q = list(map(int, input().split()))
MOD = 10 ** 9 + 7
INV2 = (MOD + 1) // 2
lines = sys.stdin.readlines()
aaa = list(map(int, lines[:n]))
mat = [[0] * n for _ in [0] * n]
for i in range(n):
    for j in range(n):
        mat[i][j] = int(aaa[i] < aaa[j])
# print(*mat, sep='\n')
for line in lines[n:]:
    x, y = map(int, line.split())
    x -= 1
    y -= 1
    mat[x][y] = mat[y][x] = (mat[x][y] + mat[y][x]) * INV2 % MOD
    for i in range(n):
        if i == x or i == y:
            continue
        mat[x][i] = mat[y][i] = (mat[x][i] + mat[y][i]) * INV2 % MOD
        mat[i][x] = mat[i][y] = (mat[i][x] + mat[i][y]) * INV2 % MOD
    # print(x, y)
    # print(*mat, sep='\n')

ans = sum(sum(row[:i]) for i, row in enumerate(mat)) % MOD
ans = (ans << q) % MOD
print(ans)

```