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<bits/stdc++.h>
using namespace std;
const int N=110;
int n,b[N],x[N],y[N],s[N];
int main(){
	scanf("%d",&n);
	for(int i=1; i<=n; ++i)
		scanf("%d",&b[i]);
	for(int i=1; i<=n; ++i)x[i]=i,y[i]=1;
	for(int i=1; i<=n; ++i){
		int k=0;
		for(int j=1; j<=n; ++j)if(b[j]==x[j]&&y[j]==1)k=j;
		if(k==0){puts("-1"); return 0;}
		for(int j=k; j<=n; ++j)--x[j]; y[k]=0; s[++s[0]]=b[k];
	}
	for(int i=n; i>=1; --i)printf("%d\n",s[i]);
	return 0;
}