容易得出数组中具有单调性:随着k范围增大，多数相与运算的结果单调非减 因此具有单调性
可以二分 
复杂度O(20nlogn)
code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10, bits = 20;
int t;
int pre[N][bits];
int a[N];
int n;

void init(){
    for(int i = 1;i <= n;i++){
        for(int j = 0;j < bits;j++){
            if(a[i] & (1 << j)){
                pre[i][j] = pre[i - 1][j] + 1;
            }
            else{
                pre[i][j] = pre[i - 1][j];
            }
        }
    }
}

bool check(int x){
    int last;
    for(int i = x;i <= n;i++){
        ll ans = 0;
        for(int j = 0;j < bits;j++){
            if(pre[i][j] - pre[i - x][j])ans += 1 << j;
        }
        // cout << ans << endl;
        if(i == x){
            last = ans;
        }
        else{
            if(ans != last)return false;
        }
    }
    return true;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while(t--){
        cin >> n;
        for(int i = 1;i <= n;i++)cin >> a[i];
        init();
        int l = 1, r = n;
        while(l < r){
            int mid = (l + r) >> 1;
            // cout << mid << endl;
            if(check(mid))r = mid;
            else l = mid + 1;
        }
        cout << l << endl;
    }
    return 0;
}