#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int rund() {
  int x576363482791fuweh = rng();
  return abs(x576363482791fuweh) % RAND_MAX;
}
const char en = '\n';
vector<int> bases;
ll myrand() {
  ll a = rund();
  a *= RAND_MAX;
  a += rund();
  return a;
}
const ll llinf = 1LL << 60;
const int inf = 1 << 30;
const int MOD = 1e9 + 7;
void yes() {
  cout << "YES" << en;
  return;
}
void no() {
  cout << "NO" << en;
  return;
}
inline int pot(int n, int k) {
  ll ret = 1;
  while (k) {
    if (k & 1) {
      ret *= n;
      ret %= MOD;
    }
    n *= n;
    n %= MOD;
    k >>= 1;
  }
  return ret;
}
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
const int MAXN = 1e5 + 10;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (int i = 0; i < 10; ++i)
    bases.push_back(rand() % (MOD - 2 * 13893829) + 13893829);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v(n);
    for (auto& it : v) cin >> it;
    vector<int> ls(n + 1, -1);
    int sol = inf;
    for (int i = 0; i < n; ++i) {
      if (ls[v[i]] != -1) sol = min(sol, i - ls[v[i]] + 1);
      ls[v[i]] = i;
    }
    if (sol != inf)
      cout << sol;
    else
      cout << -1;
    cout << en;
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int rund() {
  int x576363482791fuweh = rng();
  return abs(x576363482791fuweh) % RAND_MAX;
}
const char en = '\n';
vector<int> bases;
ll myrand() {
  ll a = rund();
  a *= RAND_MAX;
  a += rund();
  return a;
}
const ll llinf = 1LL << 60;
const int inf = 1 << 30;
const int MOD = 1e9 + 7;
void yes() {
  cout << "YES" << en;
  return;
}
void no() {
  cout << "NO" << en;
  return;
}
inline int pot(int n, int k) {
  ll ret = 1;
  while (k) {
    if (k & 1) {
      ret *= n;
      ret %= MOD;
    }
    n *= n;
    n %= MOD;
    k >>= 1;
  }
  return ret;
}
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
const int MAXN = 1e5 + 10;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (int i = 0; i < 10; ++i)
    bases.push_back(rand() % (MOD - 2 * 13893829) + 13893829);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v(n);
    for (auto& it : v) cin >> it;
    vector<int> ls(n + 1, -1);
    int sol = inf;
    for (int i = 0; i < n; ++i) {
      if (ls[v[i]] != -1) sol = min(sol, i - ls[v[i]] + 1);
      ls[v[i]] = i;
    }
    if (sol != inf)
      cout << sol;
    else
      cout << -1;
    cout << en;
  }
  return 0;
}

```