很有意思的一道题。要求分组之后的和最小，同时，分组的个数要尽量多。首先发现如果某一组的强度是 $0$，那么所有组的强度之后最坏应该是 $0$（因为此时所有元素的按位与一定是 $0$）。又要求分组尽量多，所以就需要找到尽量多的子数组，使得这个子数组的按位与是 $0$，对于其中剩下的那些按位与不是 $0$ 的元素，把它们分到相邻的任意一个合法的组即可。如果找不到任何一个组的强度是 $0$，那么最终的解是整个数组，因为任何一个子数组的强度都是正数，分开之后强度和会更大。

// Date: Sat Dec  9 20:25:47 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 200010;
int t, a[N], n;

int main(void) {
#ifdef _DEBUG
  freopen("1847b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n;

    For1(i, 1, n) { cin >> a[i]; }

    int res = 0;
    ll cur = 0;
    bool flag = true;

    For1(i, 1, n) {
      if (flag) {
        cur = a[i];
        flag = false;
      } else {
        cur = cur & a[i];
      }

      if (cur == 0) {
        res++;
        flag = true;
      }
    }

    if (!res)
      res = 1;

    cout << res << '\n';
  }

  return 0;
}