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<bits/stdc++.h>
using namespace std;
const int N=5e5+50,mod=1e9+7;
int n,m,a[N],f[N],s1[N],s2[N],J[N],ans,cnt,dat,sum;bool flag[N];
void add(int x){for(;x<=n;x+=x&-x)f[x]++;}
int ask(int x){int y=0;for(;x;x-=x&-x)y+=f[x];return y;}
int main(){
    scanf("%d",&n);J[0]=1;
    for(int i=1;i<=n;i++)J[i]=1ll*J[i-1]*i%mod;
    for(int i=1;i<=n;i++)scanf("%d",&a[i]),flag[a[i]]=1;
    for(int i=1;i<=n;i++)if(!flag[i])(sum+=i-1)%=mod;
    for(int i=1;i<=n;i++)s1[i]=s1[i-1]+!flag[i];
    for(int i=n;i;i--)s2[i]=s2[i+1]+!flag[i];m=s1[n];
    for(int i=1;i<=n;i++){
        int anss=0;
        if(a[i]){
            add(a[i]);
            anss=(anss+1ll*(a[i]-ask(a[i]))*J[m])%mod;
            if(m)anss=(anss-1ll*s1[a[i]]*cnt%mod*J[m-1])%mod;
            (dat+=s2[a[i]])%=mod;
        }
        else{
            anss=(anss+1ll*(sum-dat)*J[m-1])%mod;
            if(m>1)anss=(anss-1ll*m*(m-1)/2%mod*cnt%mod*J[m-2])%mod;
            cnt++;
        }
        ans=(ans+1ll*anss*J[n-i])%mod;
    }
    printf("%d\n",((ans+J[m])%mod+mod)%mod);
    return 0;
}