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.
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long ar[35]{};
    long long l, r;
    cin >> l >> r;
    if (l % 2 == 1 && r % 2 == 1) {
      ar[0] = (r - l) / 2 + 1;
    } else if (l % 2 == 1 && r % 2 == 0) {
      ar[0] = (r - l + 1) / 2;
    } else if (l % 2 == 0 && r % 2 == 0) {
      ar[0] = (r - l) / 2;
    } else if (l % 2 == 0 && r % 2 == 1) {
      ar[0] = (r - l + 1) / 2;
    }
    for (long long i = 1; i < 30; i++) {
      long long p = (1 << i);
      long long q = (1 << (i + 1));
      if (l <= p) {
        if (r >= q) {
          ar[i] = q - p;
        } else {
          ar[i] = q - p - (q - r - 1);
        }
      } else {
        if (r >= q) {
          ar[i] = q - p - (l - p);
        } else {
          ar[i] = q - p - (l - p + q - r) + 1;
        }
      }
    }
    long long mx = *max_element(ar, ar + 35);
    cout << r - l + 1 - mx << '\n';
  }
}
