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;
long long int arr[100001];
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int a, b;
    cin >> a >> b;
    if ((int)log2(b) - (int)log2(a) == 0) {
      cout << "0" << endl;
    } else {
      if ((int)log2(b) - (int)log2(a) == 1) {
        int f = log2(b);
        long long int x = pow(2, f);
        int xf = b - a + 1;
        if (xf % 2 == 1) {
          if (a % 2 == 1 or b % 2 == 1) {
            xf = xf / 2 + 1;
          } else {
            xf = xf / 2;
          }
        } else {
          xf = xf / 2;
        }
        cout << min(min((b - a + 1) - (x - a), b - a + 1 - xf),
                    (b - a + 1) - (b - x))
             << endl;
      } else if ((int)log2(b) - (int)log2(a) >= 2) {
        int x = b - pow(2, log2(b)) + 1;
        int y = pow(2, log2(b)) - pow(2, log2(b) - 1);
        int f = b - a + 1;
        if (f % 2 == 1) {
          if (a % 2 == 1 or b % 2 == 1) {
            f = f / 2 + 1;
          } else {
            f = f / 2;
          }
        } else {
          f = f / 2;
        }
        cout << min(b - a + 1 - x, min(b - a + 1 - y, b - a + 1 - f)) << endl;
      }
    }
  }
  return 0;
}
