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 solver() {
  int l, r;
  cin >> l >> r;
  if (l == r) {
    return 0;
  }
  int ans = INT_MAX;
  for (int i = 1; i <= 18; ++i) {
    int m = 0;
    for (int j = 18; j > i; --j) {
      if ((l & (1 << j)) > 0) {
        m = m | (1 << j);
      }
    }
    int t = 1 << (i + 1);
    int tp = 1 << i;
    int d = r - m + 1;
    int c = (d / t) * tp + min(d % t, tp);
    int d2 = l - m;
    int c2 = (d2 / t) * tp + min(d2 % t, tp);
    ans = min(ans, c - c2);
  }
  return ans;
}
int main() {
  int q = 1;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cout << solver();
    cout << '\n';
  }
}
