1257_C. Dominated Subarray
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.
```python
import collections
ii = lambda: int(input())
iia = lambda: list(map(int,input().split()))
isa = lambda: list(input().split())

t = ii()

for i in range(t):
	n = ii()
	x = iia()
	if(n==1):
		print('-1')
	else:
		dist = collections.defaultdict(lambda : n)
		last = collections.defaultdict(lambda : -1)
		mini = n
		flag = 0
		for i in range(n):
			if(dist[x[i]] == n and last[x[i]] == -1):
				last[x[i]] = i
			elif(last[x[i]] != -1 and dist[x[i]] > i-last[x[i]]):
				dist[x[i]] = i-last[x[i]]+1
				last[x[i]] = i
				mini = min(dist[x[i]], mini)
				flag = 1
		if(flag == 0):
			print(-1)
		else:
			print(mini)

```