#include <bits/stdc++.h>
#pragma GCC optimize("-O3")
using namespace std;
const long long N = 1e6 + 5;
const long long MAX = 1e7 + 9;
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long isprime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  for (long long i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
long long mod = 1e9 + 7;
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % mod;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % mod;
    res %= mod;
  }
  return res;
}
long long countDistinct(long long arr[], long long n) {
  sort(arr, arr + n);
  long long res = 0;
  for (long long i = 0; i < n; i++) {
    while (i < n - 1 && arr[i] == arr[i + 1]) i++;
    res++;
  }
  return res;
}
long long int2Bin(string s) {
  reverse(s.begin(), s.end());
  long long b = 1;
  long long x = 0;
  for (long long i = 0; i < s.length(); ++i) {
    if (s[i] == '1') x += b;
    b *= 2;
    ;
  }
  return x;
}
long long getSum(long long n) {
  long long sum;
  for (sum = 0; n > 0; sum += n % 10, n /= 10)
    ;
  return sum;
}
long long nex(long long i) { return (i / 10 + 1) * 10 - i; }
bool up[N];
long long a[N], ind, leave[11];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    pair<long long, long long> a[n];
    long long i;
    for (i = 0; i < n; i++) {
      long long x;
      cin >> x;
      a[i].first = x;
      a[i].second = i;
    }
    sort(a, a + n);
    long long ans = 2 * n;
    for (i = 1; i < n; i++) {
      if (a[i].first == a[i - 1].first)
        ans = min(ans, a[i].second - a[i - 1].second + 1);
    }
    if (ans == 2 * n)
      cout << "-1\n";
    else
      cout << ans << "\n";
  }
}

### Prompt

Your task is to create 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>
#pragma GCC optimize("-O3")
using namespace std;
const long long N = 1e6 + 5;
const long long MAX = 1e7 + 9;
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long isprime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  for (long long i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
long long mod = 1e9 + 7;
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % mod;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % mod;
    res %= mod;
  }
  return res;
}
long long countDistinct(long long arr[], long long n) {
  sort(arr, arr + n);
  long long res = 0;
  for (long long i = 0; i < n; i++) {
    while (i < n - 1 && arr[i] == arr[i + 1]) i++;
    res++;
  }
  return res;
}
long long int2Bin(string s) {
  reverse(s.begin(), s.end());
  long long b = 1;
  long long x = 0;
  for (long long i = 0; i < s.length(); ++i) {
    if (s[i] == '1') x += b;
    b *= 2;
    ;
  }
  return x;
}
long long getSum(long long n) {
  long long sum;
  for (sum = 0; n > 0; sum += n % 10, n /= 10)
    ;
  return sum;
}
long long nex(long long i) { return (i / 10 + 1) * 10 - i; }
bool up[N];
long long a[N], ind, leave[11];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    pair<long long, long long> a[n];
    long long i;
    for (i = 0; i < n; i++) {
      long long x;
      cin >> x;
      a[i].first = x;
      a[i].second = i;
    }
    sort(a, a + n);
    long long ans = 2 * n;
    for (i = 1; i < n; i++) {
      if (a[i].first == a[i - 1].first)
        ans = min(ans, a[i].second - a[i - 1].second + 1);
    }
    if (ans == 2 * n)
      cout << "-1\n";
    else
      cout << ans << "\n";
  }
}

```