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>
#include <cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
#include<complex>
using namespace std;

#define rep(i,x) for(ll i=0;i<x;i++)
#define repn(i,x) for(ll i=1;i<=x;i++)

typedef long long ll;
const ll INF = 1e18+1e18;
const ll MOD = 1000000007;
const ll MAX = 4000001;
const long double eps = 1E-14;

ll max(ll a, ll b) {
	if (a > b) { return a; }
	return b;
}

ll min(ll a, ll b) {
	if (a > b) { return b; }
	return a;
}

ll gcd(ll a, ll b) {
	if (b == 0) { return a; }
	if (a < b) { return gcd(b, a); }
	return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

struct edge {
	ll ind;
	ll fr;
	ll to;
	ll d;
};

class mint {
	long long x;
public:
	mint(long long x = 0) : x((x% MOD + MOD) % MOD) {}
	mint operator-() const {
		return mint(-x);
	}
	mint& operator+=(const mint& a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint& a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const  mint& a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint& a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint& a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint& a) const {
		mint res(*this);
		return res *= a;
	}

	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint& a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint& a) const {
		mint res(*this);
		return res /= a;
	}

	friend ostream& operator<<(ostream& os, const mint& m) {
		os << m.x;
		return os;
	}
};

mint pw(mint a, ll b) {
	if (b == 0) { return 1; }
	mint ret = pw(a, b >> 1);
	ret *= ret;
	if (b & 1) { ret *= a; }
	return ret;
}

typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<vector<ll>>> vvvll;

typedef vector<mint> vmint;
typedef vector<vector<mint>> vvmint;
typedef vector<vector<vector<mint>>> vvvmint;

//////////////////////////////////////
template<typename T>
struct Segment_Tree {
	typedef function<T(T, T)> F;

	ll n;
	F f;//モノイドとしての演算(最小値、和　など)
	T unit;//単位元

	ll size;
	vector<T> seg;

	Segment_Tree(ll n0, F f0, T unit0) {
		n = n0;
		f = f0;
		unit = unit0;

		size = 1;
		while (size < n) {
			size *= 2;
		}
		seg.assign(2 * size, unit);
	}


	void update(ll k, T x) {//k番目の要素をxに更新
		k += size - 1;
		seg[k] = x;
		while (k > 1) {
			k /= 2;
			seg[k] = f(seg[2 * k], seg[2 * k + 1]);
		}
	}



	T query_prep(ll a, ll b, ll k, ll l, ll r) {//[a,b)とkに対応する区間[l,r)の共通部分でクエリ
		if (b <= l || a >= r) { return unit; }
		if (a <= l && r <= b) { return seg[k]; }

		T v1 = query_prep(a, b, 2 * k, l, (l + r) / 2);
		T v2 = query_prep(a, b, 2 * k + 1, (l + r) / 2, r);
		return f(v1, v2);

	}

	T query(ll a, ll b) {
		return query_prep(a, b, 1, 1, size + 1);
	}

};



int main() {
	ll N;
	cin >> N;
	vll p(N + 1);
	repn(i, N)cin >> p[i];

	ll M = 0;//0の個数
	vll c(N + 1, 0);
	repn(i, N) {
		if (p[i] != 0) { c[p[i]] = i; }
		else { M++; }
	}

	//repn(i, N) { cout << c[i] << " "; }
	//cout << endl;

	vector<ll> tento(N + 1, 0);
	Segment_Tree<ll> sg(N, [](ll a, ll b) {return a + b; }, 0);
	repn(i, N) {
		if (c[i] > 0) { sg.update(c[i], 1); }
		tento[c[i]] = sg.query(c[i] + 1, N + 1);
	}

	//repn(i, N) { cout << tento[i] << " "; }
	//cout << endl;

	vll miman(N + 1, 0);//自分未満の残り物の数
	repn(i, N) {
		if (i > 1 && c[i - 1] == 0) { miman[i] = miman[i - 1] + 1; }
		else { miman[i] = miman[i - 1]; }
	}
	
	//repn(i, N) { cout << miman[i] << " "; }
	//cout << endl;

	vll znum(N + 1, 0);
	for (ll i = N - 1; i >= 1; i--) {
		if (p[i + 1] == 0) { znum[i] = znum[i + 1] + 1; }
		else { znum[i] = znum[i + 1]; }
	}

	//repn(i, N) { cout << znum[i] << " "; }
	//cout << endl;

	vll rui(N + 1, 0);
	for (ll i = N - 1; i >= 1; i--) {
		if (p[i + 1] != 0) { rui[i] = rui[i + 1] + M - miman[p[i + 1]]; }
		else { rui[i] = rui[i + 1]; }
	}

	//repn(i, N) { cout << rui[i] << " "; }
	//cout << endl;

	vmint frac(N + 1, 1);
	repn(i, N)frac[i] = frac[i - 1] * i;

	mint ans = 0;
	repn(i, N) {
		mint now = 0;

		if (p[i] != 0) {
			now += frac[M] * tento[i];
			if (M > 0)now += frac[M - 1] * miman[p[i]] * znum[i];
		}

		else {
			now += frac[M - 1] * rui[i];
			now += frac[M] / 2 * znum[i];
		}

		now *= frac[N - i];
		ans += now;
	}

	ans += frac[M];
	cout << ans << endl;




}