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;
  const long long mod = 1e9 + 7;
  const int MAX = 1e9 + 1;
  const long long modd = 998244353;
  const int MAXN = 2e5 + 10;
  const long long val = 1073741824;
  long long cnt = 0;
  vector<int> prime(MAXN, 0);
  vector<int> check(MAXN, 0);
  long long inv(long long i) {
    if (i == 1) return 1;
    return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
  }
  long long mod_mul(long long a, long long b) {
    a = a % mod;
    b = b % mod;
    return (((a * b) % mod) + mod) % mod;
  }
  long long mod_add(long long a, long long b) {
    a = a % mod;
    b = b % mod;
    return (((a + b) % mod) + mod) % mod;
  }
  long long gcd(long long a, long long b) {
    if (b > a) swap(a, b);
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long ceil_div(long long a, long long b) {
    return a % b == 0 ? a / b : a / b + 1;
  }
  long long pwr(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a;
      a = a * a;
      b >>= 1;
    }
    return res;
  }
  long long pwr2(long long a, long long b) {
    if (b == 0) return 1;
    if (b == 1) return a;
    long long ans = pwr2(a, b / 2);
    if (b % 2 == 0) return (ans * ans) % modd;
    return ((ans * ans) % modd * a) % modd;
  }
  void traversal(vector<vector<int>> &mat, int n, int m, int i, int j, int &cnt) {
    if (i >= n || i < 0 || j >= m || j < 0 || mat[i][j] != 1) {
      return;
    }
    mat[i][j] = 2;
    cnt++;
    traversal(mat, n, m, i + 1, j, cnt);
    traversal(mat, n, m, i - 1, j, cnt);
    traversal(mat, n, m, i, j + 1, cnt);
    traversal(mat, n, m, i, j - 1, cnt);
  }
  static bool compare(pair<int, int> a, pair<int, int> b) {
    if (a.first == b.first) {
      return a.second < b.second;
    }
    return a.first > b.first;
  }
  void generate_prime() {
    for (int i = 2; i <= 10000000; i++) {
      if (!check[i]) prime[cnt++] = i;
      for (int j = 0; j < cnt && prime[j] * i <= 10000000; j++) {
        check[i * prime[j]] = 1;
        if (i % prime[j] == 0) break;
      }
    }
  }
  int gcd(int a, int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  int lcm(int a, int b) { return (a * b) / gcd(a, b); }
  int nCr(int n, int r) {
    int p = 1, k = 1;
    if (n - r < r) r = n - r;
    if (r != 0) {
      while (r) {
        p *= n;
        k *= r;
        int m = gcd(p, k);
        p /= m;
        k /= m;
        n--;
        r--;
      }
    } else {
      p = 1;
    }
    return p;
  }
  long long rev(long long x) {
    long long ans = 0, m = 0;
    m = log2(x) + 1;
    for (int i = 0; i < m; i++) {
      if (x & (1LL << i)) ans += (1LL << (m - i - 1));
    }
    return ans;
  }
  int pre[MAXN][21];
  void precalculate() {
    memset(pre, 0, sizeof(pre));
    for (int i = 1; i < MAXN; i++) {
      for (int j = 0; j <= 20; j++) {
        if (i < (1 << j)) break;
        if (i & (1 << j)) {
          pre[i][j] = 1;
        }
        pre[i][j] += pre[i - 1][j];
      }
    }
  }
  int solve() {
    long long t, i = 0, j = 0, sum = 0, k = 0, l, r, temp, count = 0;
    long long n, m, c = 0, nH = 0, nT = 0, x, y, z, ans = INT_MIN, zero = 0,
                    one = 0;
    cin >> l >> r;
    n = log2(l);
    m = log2(r);
    if (n == m) {
      cout << 0 << '\n';
      return 0;
    }
    for (int i = 0; i <= 20; i++) {
      temp = pre[r][i] - pre[l - 1][i];
      ans = max(ans, temp);
    }
    cout << r - l + 1 - ans << '\n';
    return 0;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t, n, i, j, ans, temp, sum, k;
    string sans;
    t = 1;
    precalculate();
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  