Given an array a of length n, find another array, b, of length n such that:

  * for each i (1 ≤ i ≤ n) MEX(\\{b_1, b_2, …, b_i\})=a_i. 



The MEX of a set of integers is the smallest non-negative integer that doesn't belong to this set.

If such array doesn't exist, determine this.

Input

The first line contains an integer n (1 ≤ n ≤ 10^5) — the length of the array a.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ i) — the elements of the array a. It's guaranteed that a_i ≤ a_{i+1} for 1≤ i < n.

Output

If there's no such array, print a single line containing -1.

Otherwise, print a single line containing n integers b_1, b_2, …, b_n (0 ≤ b_i ≤ 10^6)

If there are multiple answers, print any.

Examples

Input


3
1 2 3


Output


0 1 2 

Input


4
0 0 0 2


Output


1 3 4 0 

Input


3
1 1 3


Output


0 2 1 

Note

In the second test case, other answers like [1,1,1,0], for example, are valid.
// var lineIdx = 0;
// var readline = () => {
//   var input = ["3", "1 1 3"]
//   return input[lineIdx++];
// };
// var print = (str) => {
//   console.log(str);
// };

var t = readline().split(" ").map(function (x) { return parseInt(x); })[0];

var row = readline().split(" ").map(function (x) { return parseInt(x); });

var source = new Array();
var res = new Array();

var idx = 0;

for(var i = 0; i<= t*2; i++){
  if(idx < row.length && row[idx] == i){
    while(row[idx] == i && idx < row.length) {
      idx ++;
    }
  } else{
    source.push(i)
  }
}

var last = -1;
idx = 0;
for(var j = 0; j < row.length; j++){
  if(j != 0 && row[j] != last){
    res.push(last)
  } else {
    res.push(source[idx++])
  }
  last = row[j]
}

print(res.join(' '))