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>
  const int INF = 0x3f3f3f3f;
  const int N = 2e5 + 5;
  const int mo = 1e9 + 7;
  using namespace std;
  std::vector<std::vector<int>> sum(N, std::vector<int>(22, 0));
  template <typename T>
  void read(T &x) {
    x = 0;
    int f = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar())
      if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
    x *= f;
  }
  template <typename T>
  void write(T x) {
    if (x < 0) x = -x, putchar('-');
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
  }
  template <typename T>
  void writeln(T x) {
    write(x);
    puts("");
  }
  void Pre() {
    for (int i = 1; i <= N - 5; i++)
      for (int j = 0; j <= 20; j++) sum[i][j] = sum[i - 1][j] + ((i >> j) & 1);
  }
  int main() {
    int T, l, r;
    read(T);
    Pre();
    while (T--) {
      read(l), read(r);
      int ans = 0;
      l--;
      for (int i = 0; i <= 20; i++) ans = max(ans, sum[r][i] - sum[l][i]);
      writeln(r - l - ans);
    }
  }
  