Problem: Permutation p is a sequence of integers p=[p_1, p_2, ..., 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,...,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 ≤ t ≤ 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 ≤ n ≤ 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, ..., q_n (1 ≤ q_i ≤ n) — elements of the array q for secret permutation. It is guaranteed that q_i ≤ q_{i+1} for all i (1 ≤ 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, ..., p_n (1 ≤ p_i ≤ 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
const int maxn = 1e5 + 10;
const long long linf = 1e18;
const int inf = 0x3f3f3f3f;
long long qpow(long long a, long long b) {
  return b ? ((b & 1) ? a * qpow(a * a % MOD, b >> 1) % MOD
                      : qpow(a * a % MOD, b >> 1)) %
                 MOD
           : 1;
}
long long qpow(long long a, long long b, long long c) {
  return b ? ((b & 1) ? a * qpow(a * a % MOD, b >> 1) % c
                      : qpow(a * a % MOD, b >> 1)) %
                 c
           : 1;
}
inline int ird() {
  register int x = 0, f = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * f;
}
inline long long lrd() {
  register long long x = 0, f = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * f;
}
void puts1() { cout << "-1" << endl; }
void puts2() { cout << "YES" << endl; }
void puts3() { cout << "NO" << endl; }
int gcd(int a, int b) { return (b > 0) ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a * b / gcd(a, b); }
long long lgcd(long long a, long long b) { return (b > 0) ? gcd(b, a % b) : a; }
long long llcm(long long a, long long b) { return a * b / lgcd(a, b); }
long long fac[maxn];
void initfac() {
  fac[0] = 1;
  for (long long i = 1; i <= 200000; i++) {
    fac[i] = fac[i - 1] * i % MOD;
  }
  return;
}
long long cm(long long n, long long m) {
  return (fac[n] * qpow(fac[m], MOD - 2) % MOD * qpow(fac[n - m], MOD - 2) %
          MOD) %
         MOD;
}
long long a[50];
bool cmp(long long a, long long b) { return a > b; }
int ans[maxn];
int main() {
  int t = ird();
  while (t--) {
    int n = ird();
    map<int, int> mp;
    int f = 1;
    int maxx = 0;
    for (int i = 1; i <= n; i++) {
      ans[i] = ird();
      maxx = max(maxx, ans[i]);
      if (ans[i] < maxx) {
        f = 0;
      }
      if (ans[i] < i) {
        f = 0;
      }
      if (mp[ans[i]]) {
        ans[i] = -1;
      }
      mp[ans[i]] = 1;
    }
    if (f == 0) {
      puts1();
      continue;
    }
    int j = 1;
    queue<int> q;
    while (j <= n) {
      if (!mp[j]) {
        q.push(j);
      }
      j++;
    }
    for (int i = 1; i <= n; i++) {
      if (ans[i] == -1) {
        if (!q.empty() && q.front() < ans[i - 1]) {
          while (!q.empty() && i <= n && ans[i] == -1) {
            ans[i] = q.front();
            q.pop();
            i++;
          }
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
  return 0;
}