Problem:
Score : 1200 points

Problem Statement

One day Mr. Takahashi picked up a dictionary containing all of the N! permutations of integers 1 through N.
The dictionary has N! pages, and page i (1 ≤ i ≤ N!) contains the i-th permutation in the lexicographical order.

Mr. Takahashi wanted to look up a certain permutation of length N in this dictionary, but he forgot some part of it.

His memory of the permutation is described by a sequence P_1, P_2, ..., P_N.
If P_i = 0, it means that he forgot the i-th element of the permutation; otherwise, it means that he remembered the i-th element of the permutation and it is P_i.

He decided to look up all the possible permutations in the dictionary.
Compute the sum of the page numbers of the pages he has to check, modulo 10^9 + 7.

Constraints

1 ≤ N ≤ 500000

0 ≤ P_i ≤ N

P_i ≠ P_j if i ≠ j (1 ≤ i, j ≤ N), P_i ≠ 0 and P_j ≠ 0.

Partial Score

In test cases worth 500 points, 1 ≤ N ≤ 3000.

Input

The input is given from Standard Input in the following format:

N
P_1 P_2 ... P_N

Output

Print the sum of the page numbers of the pages he has to check, as modulo 10^9 + 7.

Sample Input 1

4
0 2 3 0

Sample Output 1

23

The possible permutations are [1, 2, 3, 4] and [4, 2, 3, 1].
Since the former is on page 1 and the latter is on page 22, the answer is 23.

Sample Input 2

3
0 0 0

Sample Output 2

21

Since all permutations of length 3 are possible, the answer is 1 + 2 + 3 + 4 + 5 + 6 = 21.

Sample Input 3

5
1 2 3 5 4

Sample Output 3

2

Mr. Takahashi remembered all the elements of the permutation.

Sample Input 4

1
0

Sample Output 4

1

The dictionary consists of one page.

Sample Input 5

10
0 3 0 0 1 0 4 0 0 0

Sample Output 5

953330050

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

def page n, pp
	k = 1
	u = [0]
	pp[0..-2].each_with_index do |p, i|
		t = p - u.count {|u| u<p}
		b = (1...n-i).inject(:*)
		k += t * b
		u.push(p)
		# puts "[#{i}] p:#{p} t:#{t} b:#{b}"
	end
	k
end

input = $stdin.read.split("\n")
n  = input.shift.to_i
ps = input.shift.split(" ").map(&:to_i)


unused = (1..n).to_a - ps
candidates = unused.permutation.map do |us|
	ps.map do |p|
		if p==0
			us.shift
		else
			p
		end
	end
end

p candidates.map {|c| page(n, c)}.inject {|a,b| (a+b)%(10**9+7) }