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<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod=1000000007;
int p[505050];
int sum[505050];
int fac[505050];
int cnt[505050];
int lowbit(int x)
{
	return x&(-x);
}
int n;
void add(int x,int v)
{
	for(int i=x;i<=n;i+=lowbit(i))
		sum[i]+=v;
}
int query(int x)
{
	int ret=0;
	for(int i=x;i;i-=lowbit(i))
		ret+=sum[i];
	return ret;
}
int main()
{
	int le=0,ans=0,le1=0,h=0;
	scanf("%d",&n);
	fac[0]=1;
	for(int i=1;i<=n;i++)
		fac[i]=(long long)fac[i-1]*i%mod;
	for(int i=1;i<=n;i++)
		cnt[i]=1;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&p[i]);
		cnt[p[i]]--;
	}
	for(int i=n-1;i>=1;i--)
		cnt[i]+=cnt[i+1];
	for(int i=1;i<=n;i++)
		if(p[i]==0)
			le++;
	memset(sum,0,sizeof(sum));
	ans=fac[le];
	for(int i=n;i>=1;i--)
	{
		if(p[i])
		{
			ans=((long long)query(p[i])*fac[le]%mod*fac[n-i]+ans)%mod;
			ans=((long long)(cnt[1]-cnt[p[i]])*le1%mod*fac[le-1]%mod*fac[n-i]+ans)%mod;
			add(p[i],1);
			(h+=cnt[p[i]])%=mod;
		}
		else
		{
			if(le>1)
				ans=((long long)le*(le-1)/2%mod*le1%mod*fac[le-2]%mod*fac[n-i]+ans)%mod;
			ans=((long long)h*fac[le-1]%mod*fac[n-i]+ans)%mod;
			le1++;
		}
	}
	printf("%d",ans);
	return 0;
}