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: Runtime Error)

#include"bits/stdc++.h"
using namespace std;

//typedef
//------------------------------------------
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;
typedef vector<string> VS;
typedef pair<LL, LL> PLL;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define PB push_back
#define MP make_pair
#define SORT(c) sort((c).begin(),(c).end())
#define COUT(x) cout<<x<<endl

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

//constant
//--------------------------------------------
const double PI = acos(-1.0);
const LL MOD = 1e9 + 7;
const ULL INF= 1e18;
const int dx[]{ 1,-1,0,0 }, dy[]{ 0,0,1,-1 };

//function
//--------------------------------------------
LL gcd(LL, LL);
bool isEven(LL);
void coutY();
void coutN();

int main() {
	LL n; cin >> n;
	VLL b(n); REP(i, n)cin >> b[i];
	VLL ans;
	for (int i = b.size()-1; i >=0&&!b.empty();) {
		if (b[i] == i + 1) {
          //cout<<b[i]<<endl;
			ans.PB(b[i]);
			b.erase(b.begin() + i);
		}
		else {
			--i;
		}
	}
	if (b.empty()) {
		reverse(ALL(ans));
		REP(i, n)COUT(ans[i]);
	}
	else {
		COUT(-1);
	}
}

LL gcd(LL a, LL b) {
	if (a == 0 || b == 0)return 0;
	if (a < b)swap(a, b);
	if (a % b == 0)return b;
	else gcd(b, a % b);
}
bool isEven(LL a) {
	return !((bool)(a % 2));
}
void coutY() {
	cout << "Yes" << endl;
	return;
}
void coutN() {
	cout << "No" << endl;
	return;
}