#include <bits/stdc++.h>
using namespace std;
template <class C>
void min_self(C &a, C b) {
  a = min(a, b);
}
template <class C>
void max_self(C &a, C b) {
  a = max(a, b);
}
long long mod(long long n, long long m = 1000000007) {
  n %= m, n += m, n %= m;
  return n;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
unordered_map<long long, int, custom_hash> safe_map;
const int MAXN = 2e5 + 5;
const int LOGN = 21;
const long long INF = 1e14;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
vector<int> ind[MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int test = 1; test < t + 1; test++) {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      ind[i].clear();
    }
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      ind[x - 1].push_back(i);
    }
    int ans = 1e8;
    for (int i = 0; i < n; i++) {
      for (int j = 1; j < ind[i].size(); j++) {
        ans = min(ans, ind[i][j] - ind[i][j - 1] + 1);
      }
    }
    if (ans == 1e8) ans = -1;
    cout << ans, cout << "\n";
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
template <class C>
void min_self(C &a, C b) {
  a = min(a, b);
}
template <class C>
void max_self(C &a, C b) {
  a = max(a, b);
}
long long mod(long long n, long long m = 1000000007) {
  n %= m, n += m, n %= m;
  return n;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
unordered_map<long long, int, custom_hash> safe_map;
const int MAXN = 2e5 + 5;
const int LOGN = 21;
const long long INF = 1e14;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
vector<int> ind[MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int test = 1; test < t + 1; test++) {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      ind[i].clear();
    }
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      ind[x - 1].push_back(i);
    }
    int ans = 1e8;
    for (int i = 0; i < n; i++) {
      for (int j = 1; j < ind[i].size(); j++) {
        ans = min(ans, ind[i][j] - ind[i][j - 1] + 1);
      }
    }
    if (ans == 1e8) ans = -1;
    cout << ans, cout << "\n";
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

```