description: You are given an array consisting of all integers from [l, r] inclusive. For example, if l = 2 and r = 5, the array would be [2, 3, 4, 5]. What's the minimum number of elements you can delete to make the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the array non-zero?
  
  A bitwise AND is a binary operation that takes two equal-length binary representations and performs the AND operation on each pair of the corresponding bits.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
  
  The first line of each test case contains two integers l and r (1 ≤ l ≤ r ≤ 2 ⋅ 10^5) — the description of the array.
  
  Output
  
  For each test case, output a single integer — the answer to the problem.
  
  Example
  
  Input
  
  
  5
  1 2
  2 8
  4 5
  1 5
  100000 200000
  
  
  Output
  
  
  1
  3
  0
  2
  31072
  
  Note
  
  In the first test case, the array is [1, 2]. Currently, the bitwise AND is 0, as 1\ \& \ 2 = 0. However, after deleting 1 (or 2), the array becomes [2] (or [1]), and the bitwise AND becomes 2 (or 1). This can be proven to be the optimal, so the answer is 1.
  
  In the second test case, the array is [2, 3, 4, 5, 6, 7, 8]. Currently, the bitwise AND is 0. However, after deleting 4, 5, and 8, the array becomes [2, 3, 6, 7], and the bitwise AND becomes 2. This can be proven to be the optimal, so the answer is 3. Note that there may be other ways to delete 3 elements.
solution: #include <bits/stdc++.h>
  using namespace std;
  long long dirs8[] = {0, -1, -1, 0, 1, -1, 1, 1, 0};
  long long dirs4[] = {0, 1, 0, -1, 0};
  long long dirs2[] = {0, 1, 0};
  long long fact(long long n) {
    long long ans = 1;
    while (n) ans *= n--;
    return ans;
  }
  bool isPrime(long long &n) {
    for (long long i = 2; i * i <= n; ++i)
      if (n % i == 0) return 0;
    return 1;
  }
  vector<pair<long long, long long>> getPrimeFactors(long long n) {
    vector<pair<long long, long long>> facts;
    for (long long i = 2; i * i <= n; ++i) {
      if (n % i == 0) {
        long long cnt = 0;
        while (n % i == 0) {
          ++cnt;
          n /= i;
        }
        facts.push_back({i, cnt});
      }
    }
    if (n > 1) facts.push_back({n, 1});
    return facts;
  }
  const long long N = 200005;
  long long dp[N][32];
  void solve() {
    long long low, high;
    cin >> low >> high;
    low--;
    long long res = INT_MAX;
    for (long long i = 0; i < 32; i++) {
      res = min(res, high - low - dp[high][i] + dp[low][i]);
    }
    cout << res << '\n';
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    memset(dp[0], 0, sizeof dp[0]);
    for (long long i = 1; i <= N; i++) {
      for (long long c = 0; c < 32; c++) {
        dp[i][c] = dp[i - 1][c];
      }
      bitset<19> b(i);
      for (long long j = 0; j < 32; j++) {
        if (b[j] == true) {
          dp[i][j]++;
        }
      }
    }
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  