Snuke has an empty sequence a.

He will perform N operations on this sequence.

In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).

You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.

Constraints

* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N

Input

Input is given from Standard Input in the following format:


N
b_1 \dots b_N


Output

If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.

Examples

Input

3
1 2 1


Output

1
1
2


Input

2
2 2


Output

-1


Input

9
1 1 1 2 2 1 2 3 2


Output

1
2
2
3
1
2
2
1
1
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)

using namespace std;
typedef long long ll;
const ll N=12e4;

int a[N];
int p[N];
int n;

int main()
{
	scanf("%d",&n);
	fo(i,1,n)scanf("%d",&a[i]);
	int ok=1;
	fd(i,n,1){
		int po=0;
		fd(l,i,1)if(a[l]==l){
			po=l;
			break;
		}
		fo(l,po,i-1)a[l]=a[l+1];
		if(!po){
			ok=0; break;
		}
		p[i]=po;
	}
	if(!ok)puts("-1");
	else
	fo(i,1,n)printf("%d\n",p[i]);
}