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 JavaScript (node.js v5.12) (judge verdict: Accepted)

"use strict";
function I(s,f){this._s=s.split("\n");this._c=0;this._l=0;this._f=f||Number}I.prototype.a=function(){var l=this._s[this._l];if(!l)return;var t=l.trim().split(" ");var a=t[this._c];this._c++;if(t.length===this._c){this._c=0;this._l++}return this._f(a)};I.prototype.l=function(){var l=this._s[this._l];if(!l)return;this._c=0;this._l++;return l.split(" ").map(this._f)};module.exports=I;

function main(input) {
  const o = new I(input);

  const N = o.a();
  let b = [];
  for (let i = 0; i < N; i++) {
    b.push(o.a());
  }

  let result = [];
  while (b.length > 0) {
    let flag = true;
    for (let i = b.length - 1; i>=0;i--) {
      if (b[i] === i + 1) {
        let removed = b.splice(i, 1);
        result.push(removed[0]);
        flag = false;
        break;
      }
    }

    if (flag) {
      break;
    }
  }

  if (b.length === 0) {
    result.reverse().forEach(r => console.log(r));
  } else {
    console.log("-1");
  }


}


main(require("fs").readFileSync("/dev/stdin", "utf8"));