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 <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <ctime>
#include <deque>
#include <unordered_set>
using namespace std;
#define N 610000
#define P 1000000007

int n, p[N], b[N], empty_a, empty, tot, sum[N], prod[N], inv[N];
int used[N];
int ans;

int calc(int x) {
	int ans = 0;
	for (; x; x -= x & -x)
		ans += b[x];
	return ans;
}

void upd(int x) {
	for (; x <= n; x += x & -x)
		b[x] += 1;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &p[i]);
	for (int i = 1; i <= n; i++)
		used[p[i]] = 1;
	for (int i = 1; i <= n; i++)
		sum[i] = sum[i - 1] + used[i];
	empty = n - sum[n];
	prod[0] = 1;
	for (int i = 1; i <= n; i++)
		prod[i] = 1LL * i * prod[i - 1] % P;
	inv[1] = 1;
	for (int i = 2; i <= n; i++)
		inv[i] = (-1LL * inv[P % i] * (P / i) % P + P) % P;

	empty_a = 0;
	tot = 0;
	for (int i = n; i; i--) {
		if (!p[i]) {
			ans = (ans + 1LL * prod[n - i] * tot % P * prod[empty - 1]) % P;
			ans = (ans + 1LL * prod[n - i] * prod[empty] % P * empty_a % P * inv[2]) % P;
			empty_a += 1;
		}else {
			ans = (ans + 1LL * prod[n - i] * calc(p[i]) % P * prod[empty]) % P;
			if (empty) {
				ans = (ans + 1LL * prod[n - i] * (p[i] - sum[p[i]]) % P * prod[empty - 1] % P * empty_a % P) % P;
			}
			upd(p[i]);
			tot += (n - p[i]) - (sum[n] - sum[p[i]]);
			tot %= P;
		}
	}
	ans = (ans + prod[empty]) % P;
	printf("%d\n", ans);
}
