Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3
1 2 1

Sample Output 1

1
1
2

In this sequence of operations, the sequence a changes as follows:

After the first operation: (1)

After the second operation: (1,1)

After the third operation: (1,2,1)

Sample Input 2

2
2 2

Sample Output 2

-1

2 cannot be inserted at the beginning of the sequence, so this is impossible.

Sample Input 3

9
1 1 1 2 2 1 2 3 2

Sample Output 3

1
2
2
3
1
2
2
1
1

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<math.h>
using namespace std;
typedef long long ll;
#define int long long
typedef vector<int> VI;
#define fore(i,a) for(auto &i:a)
#define REP(i,n) for(int i=0;i<n;i++)
#define eREP(i,n) for(int i=0;i<=n;i++)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define eFOR(i,a,b) for(int i=(a);i<=(b);++i)
#define SORT(c) sort((c).begin(),(c).end())
#define rSORT(c) sort((c).rbegin(),(c).rend())
#define LB(x,a) lower_bound((x).begin(),(x).end(),(a))
#define UB(x,a) upper_bound((x).begin(),(x).end(),(a))
#define INF 1000000000
#define LLINF 9223372036854775807
#define mod 1000000007
//vector<vector<int> > dp;
//vector<vector<vector<int> > > vvvi;
//dp=vector<vector<int> >(N, vector<int>(M,0));
//vector<pair<int,int> > v;
//v.push_back(make_pair(x,y));

int ans[110];
int N;

void dfs(VI A) {
//	cout << A.size() << endl;
//	REP(i, A.size())cout << A[i]<<" ";
//	cout << endl;
	//if (A.size() == 1 && A[0] == 1)return;
	VI B,C;

	REP(i, A.size()) {
		if (A[i] == i + 1) {
			ans[A.size()] = A[i];
			if (i == A.size() - 1) {
				dfs(B);
			}
			C.erase(C.begin(), C.end());
			REP(i, B.size())C.push_back(B[i]);

			FOR(j, i + 1, A.size()) {
				C.push_back(A[j]);

			}
			dfs(C);
		}
		B.push_back(A[i]);
	}
	if(A.size()==B.size())
		return;
			//cout << B.size() << endl;
			//REP(k, B.size())cout << B[k]<<" ";
			//cout << endl;
}

signed main(){
cin.tie(0);
ios::sync_with_stdio(false);

	
	cin >> N;
	VI A(N);
	REP(i, N) {
		cin >> A[i];
	}
	dfs(A);
	//ans[1] = 1;
	if (ans[1] == 0) {
		cout << -1 << endl;
		return 0;
	}
	FOR(i,1,N+1)cout << ans[i] << endl;

	return 0;
}