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)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double EPS = 1e-9;
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<vector<ll>> vll;
typedef pair<int, int> pi;
typedef pair<ll, ll> pll;
typedef map<int, int> mi;
typedef set<int> si;
#define VV(T) vector<vector< T > >
#define dump(x)  cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE"  ) << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible"  ) << endl

#define rep(i, n) REP(i, 0, n)                              // 0, 1, ..., n-1
#define REP(i, x, n) for(int i = x; i < n; i++)             // x, x + 1, ..., n-1
#define invrep(i, n) for(int i = (n)-1; i >= 0; i--)        // n-1, n-2, ..., 0
#define invREP(i, x, n) for(int i = (n)-1; i >= (x; i--)    // n-1, n-2, ..., x
#define FOREACH(x,a) for(auto& (x) : (a) )

#define ALL(v) (v).begin() , (v).end()
#define RALL(v) (v).rbegin(), (v).rend()

#define PB push_back

#define COUT(x) cout << (x) << endl
#define VECCIN(x) for(auto&youso_: (x) )cin>>youso_
#define VECCOUT(x) for(auto&youso_: (x) )cout<<youso_<<" ";cout<<endl

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }


int main(){
    int N; cin >> N;
    vi ans;
    rep(i, N){
        int a; cin >> a; a--;
        if(a <= i) ans.insert(ans.begin() + a, a + 1);
        else {
            cout << -1 << endl;
            return 0;
        }
    }
    for(int it: ans) printf("%d\n", it);
}