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 <iostream>
#include <string>

using namespace std;

typedef long long ll;
const int MOD = 1000000007;
int n;
int p[1 << 20];
ll fac[1 << 20];
ll st[1 << 20], tot[1 << 20];
ll bit_sum(ll st[], int x) {
  ll ans = 0;
  for (; x; x -= x & (-x)) {
    ans += st[x];
  }
  return ans;
}
void bit_add(ll st[], int x) {
  for (; x <= n; x += x & (-x)) {
    st[x]++;
  }
}

ll k = 0, sm;
ll gao(ll x, ll y, ll z) {
  ll ans = x * fac[n - k] + z * fac[n - k - 1] % MOD * y;
  return ans % MOD;
}
ll gao(ll y, ll z) {
  ll t = fac[n - k];
  if (t % 2 != 0) {
    t += MOD;
  }
  t /= 2;
  ll ans = (sm + z * t) % MOD;
  //  printf("sm=%d\n", sm);
  return ans;
}

int main() {
  scanf("%d", &n);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i % MOD;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    if (p[i]) {
      bit_add(tot, p[i]);
      k++;
    }
  }
  ll ans = fac[n - k];
  for (int i = n; i >= 1; i--) {
    if (p[i] != 0) {
      ll x = bit_sum(st, p[i]);
      ll y = p[i] - bit_sum(tot, p[i]);
      ll z = n - i - bit_sum(st, n);
      ans += gao(x, y, z) * fac[n - i];
      //      printf("++%d\n", gao(x, y, z) * fac[n - i]);
      ans %= MOD;
      bit_add(st, p[i]);
      sm += fac[n - k - 1] * (n - k - y);
      //      printf("sm+%d\n",fac[n - k - 1] * (n - k - y));
      sm %= MOD;
    } else {
      ll y = p[i] - bit_sum(tot, p[i]);
      ll z = n - i - bit_sum(st, n);
      //      printf("+%d\n", gao(y, z) * fac[n - i]);
      ans += gao(y, z) * fac[n - i];
      ans %= MOD;
    }
  }
  printf("%d\n", (int)ans);
}
