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
// ※※※ 解答不能 ※※※
// square1001氏.
// https://atcoder.jp/contests/code-festival-2016-qualc/submissions/1992158
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define repex(i, a, b, c) for(int i = a; i < b; i += c)
#define repx(i, a, b) repex(i, a, b, 1)
#define rep(i, n) repx(i, 0, n)
#define repr(i, a, b) for(int i = a; i >= b; i--)
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;
    repx(i, 2, n + 1) inv[i] = 1LL * inv[mod % i] * (mod - mod / i) % mod;
    rep(i, n) ip[i] = -1;
    repr(i, n - 1, 0){
        scanf("%d", &p[i]), p[i]--;
        if(p[i] != -1) ip[p[i]] = i;
    }
    int f1 = 1;
    rep(i, n){
        if(ip[i] == -1){
            v[c++] = i;
            f1 = 1LL * f1 * c % mod;
        }
    }
    int ret = 0, fact = 1, cur = 0, cs = 0;
    rep(i, n){
        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;
}