Permutation $p$ is a sequence of integers $p=[p_1, p_2, \dots, p_n]$, consisting of $n$ distinct (unique) positive integers between $1$ and $n$, inclusive. For example, the following sequences are permutations: $[3, 4, 1, 2]$, $[1]$, $[1, 2]$. The following sequences are not permutations: $[0]$, $[1, 2, 1]$, $[2, 3]$, $[0, 1, 2]$.

The important key is in the locked box that you need to open. To open the box you need to enter secret code. Secret code is a permutation $p$ of length $n$. 

You don't know this permutation, you only know the array $q$ of prefix maximums of this permutation. Formally:  $q_1=p_1$,  $q_2=\max(p_1, p_2)$,  $q_3=\max(p_1, p_2,p_3)$,  ...  $q_n=\max(p_1, p_2,\dots,p_n)$. 

You want to construct any possible suitable permutation (i.e. any such permutation, that calculated $q$ for this permutation is equal to the given array).


-----Input-----

The first line contains integer number $t$ ($1 \le t \le 10^4$) — the number of test cases in the input. Then $t$ test cases follow.

The first line of a test case contains one integer $n$ $(1 \le n \le 10^{5})$ — the number of elements in the secret code permutation $p$.

The second line of a test case contains $n$ integers $q_1, q_2, \dots, q_n$ $(1 \le q_i \le n)$ — elements of the array $q$ for secret permutation. It is guaranteed that $q_i \le q_{i+1}$ for all $i$ ($1 \le i < n$).

The sum of all values $n$ over all the test cases in the input doesn't exceed $10^5$.


-----Output-----

For each test case, print:  If it's impossible to find such a permutation $p$, print "-1" (without quotes).  Otherwise, print $n$ distinct integers $p_1, p_2, \dots, p_n$ ($1 \le p_i \le n$). If there are multiple possible answers, you can print any of them. 


-----Example-----
Input
4
5
1 3 4 5 5
4
1 1 3 4
2
2 2
1
1

Output
1 3 4 5 2 
-1
2 1 
1 



-----Note-----

In the first test case of the example answer $[1,3,4,5,2]$ is the only possible answer:  $q_{1} = p_{1} = 1$;  $q_{2} = \max(p_{1}, p_{2}) = 3$;  $q_{3} = \max(p_{1}, p_{2}, p_{3}) = 4$;  $q_{4} = \max(p_{1}, p_{2}, p_{3}, p_{4}) = 5$;  $q_{5} = \max(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}) = 5$. 

It can be proved that there are no answers for the second test case of the example.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'm = int(input())\nfor j in range(m):\n    n = int(input())\n    a = list(map(int, input().split()))\n    b = []\n    used = [0] * (n + 1)\n    t = 1\n    b.append(a[0])\n    f = True\n    used[a[0]] = 1\n    for k in range(1, n):\n        if a[k] == a[k - 1]:\n            while used[t] == 1:\n                t += 1                    \n            if t < a[k - 1]:\n                b.append(t)\n                used[t] = 1\n                t += 1\n            else:\n                f = False\n                break\n        elif a[k] > a[k - 1]:\n            b.append(a[k])\n            used[a[k]] = 1\n        else:\n            f = False\n            break\n    if f:\n        print(*b)\n    else:\n        print(-1)\n            \n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13443