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: Accepted)

//author Eterna
#define Hello the_cruel_world!
#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<utility>
#include<cmath>
#include<climits>
#include<deque>
#include<functional>
#include<numeric>
#include<unordered_map>
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
#define ABS(x) ((x) >= 0 ? (x) : (-(x)))
#define pb(x) push_back(x)
#define lowbit(x) (x & -x)
#define FRIN freopen("C:\\Users\\Administrator.MACHENI-KA32LTP\\Desktop\\in.txt", "r", stdin)
#define FROUT freopen("C:\\Users\\Administrator.MACHENI-KA32LTP\\Desktop\\out.txt", "w", stdout)
#define FAST ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define outd(x) printf("%d\n", x)
#define outld(x) printf("%lld\n", x)
#define il inline
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const int maxn = 100;
const int INF = 0x7fffffff;
const int mod = 1e9 + 7;
const double eps = 1e-7;
const double Pi = acos(-1.0);
inline int read_int() {
	char c;
	int ret = 0, sgn = 1;
	do { c = getchar(); } while ((c < '0' || c > '9') && c != '-');
	if (c == '-') sgn = -1; else ret = c - '0';
	while ((c = getchar()) >= '0' && c <= '9') ret = ret * 10 + (c - '0');
	return sgn * ret;
}
inline ll read_ll() {
	char c;
	ll ret = 0, sgn = 1;
	do { c = getchar(); } while ((c < '0' || c > '9') && c != '-');
	if (c == '-') sgn = -1; else ret = c - '0';
	while ((c = getchar()) >= '0' && c <= '9') ret = ret * 10 + (c - '0');
	return sgn * ret;
}
int arr[maxn + 5], n;
bool flag;
stack<int> res;
int main()
{
	n = read_int();
	for (int i = 1; i <= n; ++i)arr[i] = read_int();
	for (int i = n; i > 0; --i) {
		for (int j = i; j > 0; --j) {
			if (arr[j] > i) {
				flag = true;
				break;
			}
			if (arr[j] == j) {
				res.push(j);
				for (int k = j; k < n; ++k)arr[k] = arr[k + 1];
				break;
			}
		}
		if (flag)break;
	}
	if (flag)cout << "-1" << endl;
	else {
		for (int i = 1; i <= n; ++i) {
			printf("%d\n", res.top());
			res.pop();
		}
	}
	//system("pause");
	return 0;
}