#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    unordered_map<int, int> M;
    map<int, int> occ;
    vector<int> tab(n);
    for (int i = 0; i < n; i++) cin >> tab[i];
    if (n == 1) {
      cout << "-1" << endl;
      continue;
    }
    int ind1 = 0, ind2 = 1;
    M[tab[0]]++;
    M[tab[1]]++;
    if (tab[0] == tab[1])
      occ[2]++;
    else
      occ[1] = 2;
    int ret = n;
    bool flag = false;
    while (1) {
      if (ind1 == ind2 && ind2 == n - 1) break;
      auto it = occ.rbegin();
      if ((it->second > 1 || ind2 == ind1) && ind2 < n - 1) {
        ind2++;
        M[tab[ind2]]++;
        occ[M[tab[ind2]]]++;
        if (occ.find(M[tab[ind2]] - 1) != occ.end()) {
          occ[M[tab[ind2]] - 1]--;
          if (occ[M[tab[ind2]] - 1] == 0) occ.erase(M[tab[ind2]] - 1);
        }
      } else {
        if (it->second == 1 && ind1 < ind2) {
          flag = true;
          ret = min(ret, ind2 - ind1 + 1);
        }
        M[tab[ind1]]--;
        if (occ.find(M[tab[ind1]] + 1) != occ.end()) {
          occ[M[tab[ind1]] + 1]--;
          if (occ[M[tab[ind1]] + 1] == 0) occ.erase(M[tab[ind1]] + 1);
        }
        occ[M[tab[ind1]]]++;
        ind1++;
      }
    }
    if (flag) {
      cout << ret << endl;
    } else
      cout << "-1" << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

Let's call an array t dominated by value v in the next situation.

At first, array t should have at least 2 elements. Now, let's calculate number of occurrences of each number num in t and define it as occ(num). Then t is dominated (by v) if (and only if) occ(v) > occ(v') for any other number v'. For example, arrays [1, 2, 3, 4, 5, 2], [11, 11] and [3, 2, 3, 2, 3] are dominated (by 2, 11 and 3 respectevitely) but arrays [3], [1, 2] and [3, 3, 2, 2, 1] are not.

Small remark: since any array can be dominated only by one number, we can not specify this number and just say that array is either dominated or not.

You are given array a_1, a_2, ..., a_n. Calculate its shortest dominated subarray or say that there are no such subarrays.

The subarray of a is a contiguous part of the array a, i. e. the array a_i, a_{i + 1}, ..., a_j for some 1 ≤ i ≤ j ≤ n.

Input

The first line contains single integer T (1 ≤ T ≤ 1000) — the number of test cases. Each test case consists of two lines.

The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the array a.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the corresponding values of the array a.

It's guaranteed that the total length of all arrays in one test doesn't exceed 2 ⋅ 10^5.

Output

Print T integers — one per test case. For each test case print the only integer — the length of the shortest dominated subarray, or -1 if there are no such subarrays.

Example

Input


4
1
1
6
1 2 3 4 5 1
9
4 1 2 4 5 4 3 2 1
4
3 3 3 3


Output


-1
6
3
2

Note

In the first test case, there are no subarrays of length at least 2, so the answer is -1.

In the second test case, the whole array is dominated (by 1) and it's the only dominated subarray.

In the third test case, the subarray a_4, a_5, a_6 is the shortest dominated subarray.

In the fourth test case, all subarrays of length more than one are dominated.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    unordered_map<int, int> M;
    map<int, int> occ;
    vector<int> tab(n);
    for (int i = 0; i < n; i++) cin >> tab[i];
    if (n == 1) {
      cout << "-1" << endl;
      continue;
    }
    int ind1 = 0, ind2 = 1;
    M[tab[0]]++;
    M[tab[1]]++;
    if (tab[0] == tab[1])
      occ[2]++;
    else
      occ[1] = 2;
    int ret = n;
    bool flag = false;
    while (1) {
      if (ind1 == ind2 && ind2 == n - 1) break;
      auto it = occ.rbegin();
      if ((it->second > 1 || ind2 == ind1) && ind2 < n - 1) {
        ind2++;
        M[tab[ind2]]++;
        occ[M[tab[ind2]]]++;
        if (occ.find(M[tab[ind2]] - 1) != occ.end()) {
          occ[M[tab[ind2]] - 1]--;
          if (occ[M[tab[ind2]] - 1] == 0) occ.erase(M[tab[ind2]] - 1);
        }
      } else {
        if (it->second == 1 && ind1 < ind2) {
          flag = true;
          ret = min(ret, ind2 - ind1 + 1);
        }
        M[tab[ind1]]--;
        if (occ.find(M[tab[ind1]] + 1) != occ.end()) {
          occ[M[tab[ind1]] + 1]--;
          if (occ[M[tab[ind1]] + 1] == 0) occ.erase(M[tab[ind1]] + 1);
        }
        occ[M[tab[ind1]]]++;
        ind1++;
      }
    }
    if (flag) {
      cout << ret << endl;
    } else
      cout << "-1" << endl;
  }
  return 0;
}

```