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
const int N = 600001;
const int LG = 21;
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
const long long INFLL = 1e18;

#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <unordered_set>
#include <bitset>
#include <deque>

using namespace std;

#define int long long
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<ll> vi;

#define forn(i, n) for (int (i) = 0; (i) != (n); (i)++)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define popcount(x) __builtin_popcount(x)
#define popcountll(x) __builtin_popcountll(x)
#define fi first
#define se second
#define re return
#define uniq(x) sort(all(x)); (x).resize(unique(all(x)) - (x).begin())

#ifdef LOCAL
#define dbg(x) cerr << __LINE__ << " " << #x << " " << x << endl
#define ln cerr << __LINE__ << endl
#else
#define dbg(x) void(0)
#define ln void(0)
#endif // LOCAL

int cx[4] = {-1, 0, 1, 0};
int cy[4] = {0, -1, 0, 1};

ll inq(ll x, ll y)
{
    if (!y) re 1 % MOD;
    ll l = inq(x, y / 2);
    if (y % 2) re l * l % MOD * x % MOD;
    re l * l % MOD;
}

ll rev(ll x)
{
    return inq(x, MOD - 2);
}

bool __precomputed_combinatorics = 0;
vector<ll> __fact, __ufact, __rev;

void __precompute_combinatorics()
{
    __precomputed_combinatorics = 1;
    __fact.resize(N);
    __ufact.resize(N);
    __rev.resize(N);
    __rev[1] = 1;
    for (int i = 2; i < N; i++) __rev[i] = MOD - __rev[MOD % i] * (MOD / i) % MOD;
    __fact[0] = 1, __ufact[0] = 1;
    for (int i = 1; i < N; i++) __fact[i] = __fact[i - 1] * i % MOD, __ufact[i] = __ufact[i - 1] * __rev[i] % MOD;
}

ll fact(int x)
{
    if (!__precomputed_combinatorics) __precompute_combinatorics();
    return __fact[x];
}

ll cnk(int n, int k)
{
    if (!__precomputed_combinatorics) __precompute_combinatorics();
    return __fact[n] * __ufact[n - k] % MOD * __ufact[k] % MOD;
}

int n, p[N], fr[N], cu[N];
ll ans;

int fenw[N];
int fenw2[N];

int Get(int x)
{
    int res = 0;
    while (x >= 0)
    {
        res += fenw[x];
        x = (x & (x + 1)) - 1;
    }
    return res;
}

void Add(int x)
{
    while (x < N)
    {
        fenw[x]++;
        x = x | (x + 1);
    }
}

int Get2(int x)
{
    int res = 0;
    while (x >= 0)
    {
        res += fenw2[x];
        x = (x & (x + 1)) - 1;
    }
    return res;
}

void Add2(int x)
{
    while (x < N)
    {
        fenw2[x]++;
        x = x | (x + 1);
    }
}

signed main()
{
    srand(time(NULL));
    #ifdef ONLINE_JUDGE
        ios_base::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
    #endif // ONLINE_JUDGE
    cin >> n;
    fill(fr, fr + n, 1);
    for (int i = 0; i < n; i++)
    {
        cin >> p[i];
        p[i]--;
        fr[p[i]] = 0;
    }
    for (int i = 0; i < n; i++)
    {
        if (fr[i]) Add2(i);
    }
    cu[n - 1] = (fr[n - 1] == 1);
    for (int i = n - 2; i >= 0; i--)
    {
        cu[i] = (cu[i + 1] + (fr[i] == 1));
    }
    int k = 0;
    for (int i = 0; i < n; i++)
    {
        if (p[i] == -1) k++;
    }
    int cnt = 0;
    ll cusi = 0;
    ll si = 0;
    ll get2summ = 0;
    for (int i = 0; i < k; i++)
    {
        get2summ = (get2summ + i) % MOD;
    }
    for (int i = 0; i < n; i++)
    {
        if (fr[i] == 1) si += i;
    }
    for (int i = 0; i < n; i++)
    {
        if (p[i] != -1)
        {
            ll x = p[i];
            x = (x - Get(p[i] - 1) + MOD) % MOD;
            ans = (ans + x * fact(n - i - 1) % MOD * fact(k) % MOD) % MOD;
            ans = ((ans - fact(n - i - 1) * cnt % MOD * Get2(p[i] - 1) % MOD * fact(k - 1)) % MOD + MOD) % MOD;
            Add(p[i]);
            cusi += cu[p[i]];
        }
        else
        {
            ll s1 = ((si - cusi + MOD) % MOD + MOD) % MOD;
            ans = (ans + s1 * fact(n - i - 1) % MOD * fact(k - 1)) % MOD;
            ans = ((ans - fact(n - i - 1) * cnt % MOD * get2summ % MOD * fact(k - 2)) % MOD + MOD) % MOD;
            cnt++;
        }
    }
    //cout << ans << " " << fact(k) << endl;
    cout << (ans + fact(k)) % MOD;
}

/* Note:
Check constants at the beginning of the code.
N is set to 4e5 but be careful in problems with large constant factor.
Check corner cases.
No def int long long for now.
*/
