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.

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.

Examples

Input

4
0 2 3 0


Output

23


Input

3
0 0 0


Output

21


Input

5
1 2 3 5 4


Output

2


Input

1
0


Output

1


Input

10
0 3 0 0 1 0 4 0 0 0


Output

953330050
#include <cstdio>
#include <algorithm>
#pragma warning(disable : 4996)
using namespace std;
int n, c, p[500009], ip[500009], v[500009], bit[500009], inv[500009], mod = 1000000007;
void add(int pos, int val) {
	for (int i = pos + 1; i <= n; i += i & (-i)) bit[i] += val;
}
int sum(int pos) {
	int ret = 0;
	for (int i = pos; i >= 1; i -= i & (-i)) ret += bit[i];
	return ret;
}
int main() {
	scanf("%d", &n);
	inv[1] = 1;
	for (int i = 2; i <= n; i++) inv[i] = 1LL * inv[mod % i] * (mod - mod / i) % mod;
	for (int i = 0; i < n; i++) ip[i] = -1;
	for (int i = n - 1; i >= 0; i--) {
		scanf("%d", &p[i]), p[i]--;
		if(p[i] != -1) ip[p[i]] = i;
	}
	int f1 = 1;
	for (int i = 0; i < n; i++) {
		if (ip[i] == -1) {
			v[c++] = i;
			f1 = 1LL * f1 * c % mod;
		}
	}
	int ret = 0, fact = 1, cur = 0, cs = 0;
	for (int i = 0; i < n; i++) {
		if (p[i] != -1) {
			ret = (ret + 1LL * fact * sum(p[i]) % mod * f1) % mod;
			ret = (ret + 1LL * fact * (lower_bound(v, v + c, p[i]) - v) % mod * f1 % mod * inv[c] % mod * cur) % mod;
			cs += c - (lower_bound(v, v + c, p[i]) - v);
			if (cs >= mod) cs -= mod;
			add(p[i], 1);
		}
		else {
			ret = (ret + 1LL * fact * cs % mod * f1 % mod * inv[c]) % mod;
			ret = (ret + 1LL * fact * cur % mod * f1 % mod * inv[2]) % mod;
			cur++;
		}
		fact = 1LL * fact * (i + 1) % mod;
	}
	printf("%d\n", (ret + f1) % mod);
	return 0;
}