只需要找到一个二进制位满足所有数字对应的二进制位都为 1 即可，如果没有，那么找到二进制为 1 的数字个数最大的那个位，然后把这位上为 0 的数字删除。接下来只需要求对于一个二进制位有多少个数字满足这个位为 1，题目给定的是左右端点，所以可以前缀和优化。

// Date: Sat Nov 25 10:27:53 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, M = 32;
int t, l, r, b[N][M];

int main(void) {
#ifdef _DEBUG
  freopen("1615b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  For1(i, 1, 31) {
    For1(j, 1, int(2e5)) {
      if (j & (1 << (i - 1)))
        b[j][i] = b[j - 1][i] + 1;
      else
        b[j][i] = b[j - 1][i];
    }
  }

  cin >> t;
  while (t--) {
    cin >> l >> r;

    int res = 0, n = r - l + 1;
    For1(i, 1, 31) {
      int tmp = b[r][i] - b[l - 1][i];
      res = max(res, tmp);
    }

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

  return 0;
}