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;
int t, a, b, _a, _b, i, x, y, ans, e[200001];
int main() {
  for (i = 1; i <= 200000; i++) {
    if (i % 2 == 0)
      e[i] = e[i - 1] + 1;
    else
      e[i] = e[i - 1];
  }
  cin >> t;
  while (t--) {
    cin >> a >> b;
    ans = INT_MAX;
    if (a == b) {
      cout << 0 << endl;
      continue;
    }
    _a = log2(a), _b = log2(b);
    if (_a == _b)
      ans = 0;
    else {
      for (i = _a; i <= _b; i++) {
        x = 1 << i;
        y = 1 << (i + 1);
        y--;
        if (i == _a)
          ans = min(ans, abs(b - y));
        else if (i == _b)
          ans = min(ans, abs(x - a));
        else
          ans = min(ans, abs(a - x) + abs(b - y));
      }
    }
    cout << min(ans, e[b] - e[a - 1]) << endl;
  }
}
