Let's call an array tt dominated by value vv in the next situation.

At first, array tt should have at least 22 elements. Now, let's calculate number of occurrences of each number numnum in tt and define it as occ(num)occ(num). Then tt is dominated (by vv) if (and only if) occ(v)>occ(v′)occ(v)>occ(v′) for any other number v′v′. For example, arrays [1,2,3,4,5,2][1,2,3,4,5,2], [11,11][11,11] and [3,2,3,2,3][3,2,3,2,3] are dominated (by 22, 1111 and 33 respectevitely) but arrays [3][3], [1,2][1,2] and [3,3,2,2,1][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 a1,a2,…,ana1,a2,…,an. Calculate its shortest dominated subarray or say that there are no such subarrays.

The subarray of aa is a contiguous part of the array aa, i. e. the array ai,ai+1,…,ajai,ai+1,…,aj for some 1≤i≤j≤n1≤i≤j≤n.

The first line contains single integer TT (1≤T≤10001≤T≤1000) — the number of test cases. Each test case consists of two lines.

The first line contains single integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of the array aa.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n) — the corresponding values of the array aa.

It's guaranteed that the total length of all arrays in one test doesn't exceed 2⋅1052⋅105.

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

In the first test case, there are no subarrays of length at least 22, so the answer is −1−1.

In the second test case, the whole array is dominated (by 11) and it's the only dominated subarray.

In the third test case, the subarray a4,a5,a6a4,a5,a6 is the shortest dominated subarray.

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

#include<bits/stdc++.h>

using namespace std;

const int maxn=2e5+;

int n,a[maxn],pre[maxn];

int main() {

    int t;

    scanf("%d",&t);

    while(t--) {

        scanf("%d",&n);

        int ans=0x3f3f3f3f;

        for(int i=; i<=n; i++) pre[i]=-;

        for(int i=; i<=n; i++) {

            scanf("%d",&a[i]);//先输入

            if(pre[a[i]]!=-)//判断之前有没有出现过

                ans=min(ans,i-pre[a[i]]+);//如果出现过，取到上一个点的距离并取最小

            pre[a[i]]=i;//标记此时点的距离

        }

        printf("%d\n",ans==0x3f3f3f3f?-:ans);

    }

}

//计算每两个相同元素的距离对ans 取min