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 <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <complex>
#include <array>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<double> VD;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef vector<VD> VVD;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

template<typename T> void chmin(T &a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T &a, T b) { if (a < b) a = b; }

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }

const ll mod = 1e9 + 7;

struct BIT{
    int n;
    VL bit;

    BIT(){}
    BIT(int x){
        n = x;
        bit.assign(n+1, 0);
    }

    void init(int x){
        n = x;
        bit.assign(n+1, 0);
    }

    void add(int i, ll x){
        i++;
        while (i <= n){
            bit[i] += x;
            i += i & -i;
        }
    }

    // [0, i]
    ll sum(int i){
        i++;
        ll ret = 0;
        while (i > 0){
            ret += bit[i];
            i -= i & -i;
        }
        return ret;
    }

    // [l, r)
    ll sum(int l, int r){
        if (l >= r) return 0;
        return sum(r-1) - sum(l-1);
    }
};

ll powll(ll x, ll y){
    x %= mod;
    ll res = 1LL;
    while(y){
        if (y & 1LL)
            res *= x;
        res %= mod;
        x = (x*x) % mod;
        y >>= 1LL;
    }
    return res;
}

ll divll(ll x, ll y){
    x %= mod;
    return (x * powll(y,mod-2)) % mod;
}

int main() {
    ll n;
    cin >> n;
    BIT bit(n), bit2(n);
    REP(i,n) bit2.add(i, 1);
    VL p(n), q;
    REP(i,n){
        p[i] = in() - 1;
        if (p[i] >= 0) bit2.add(p[i], -1);
    }
    ll qs = 0;
    REP(i,n){
        if (bit2.sum(i, i + 1)){
            q.push_back(i);
            qs = (qs + i) % mod;
        }
    }
    ll m = bit2.sum(n - 1);
    VL fact(n + 1, 1);
    REP(i,n) fact[i + 1] = (fact[i] * (i + 1)) % mod;
    ll ans = 0, c = 0, del = 0, past = -1;
    REP(i,n-1){
        if (p[i] >= 0){
            ll x = p[i] - bit.sum(p[i]);
            if (m) x = (x + mod - divll(bit2.sum(p[i]) * c, m)) % mod;
            ans = (ans + x * fact[n - i - 1]) % mod;
            bit.add(p[i], 1);
            ll tmp = q.end() - lower_bound(ALL(q), p[i]);
            del = (del + tmp) % mod;
        }else{
            ll s = (qs - divll(m * c, 2) - del + 2 * mod) % mod;
            s = divll(s, m);
            ans = (ans + s * fact[n - i - 1]) % mod;
            c++;
        }
        // cout << i << " " << ans << endl;
    }
    ans = ((ans + 1) * fact[m]) % mod;
    cout << ans << endl;

    return 0;
}
