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<cstdio>
#define maxn 105
int n,a[maxn],cnt[maxn],ans[maxn];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=1;i<=n;i++) cnt[a[i]]++;
	for(int i=1;i<=n;i++)
		if(cnt[i]>n-i+1) { printf("-1\n"); return 0; }
	for(int i=n;i;i--)
	{
		int pos=0;
		for(int j=1;j<=i;j++)
			if(a[j]==j&&a[j]>a[pos]) pos=j;
		ans[i]=pos;
		for(int j=pos;j<i;j++) a[j]=a[j+1];
		for(int j=1;j<i;j++)
			if(a[j]>=i) { printf("-1\n"); return 0; }
	}
	for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
}