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;
  template <typename T>
  void rd(T& x) {
    int f = 0, c;
    while (!isdigit(c = getchar())) f ^= !(c ^ 45);
    x = (c & 15);
    while (isdigit(c = getchar())) x = x * 10 + (c & 15);
    if (f) x = -x;
  }
  template <typename T>
  void pt(T x, int c = -1) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) pt(x / 10);
    putchar(x % 10 + 48);
    if (c != -1) putchar(c);
  }
  const int N = 1000005;
  int T, n, a[N];
  int calc(int r, int b) {
    return (r >> (b + 1) << b) + (r >> b & 1 ? (r & ((1 << b) - 1)) + 1 : 0);
  }
  int main() {
    rd(T);
    while (T--) {
      int l, r;
      scanf("%d%d", &l, &r);
      int ans = 0;
      for (int i = (30); i >= (0); --i) {
        ans = max(ans, calc(r, i) - calc(l - 1, i));
      }
      printf("%d\n", r - l + 1 - ans);
    }
    return 0;
  }
  