#include <bits/stdc++.h>
using namespace std;
int f(int n, int k) {
  int r = n / k, s;
  if (r % 2) r--;
  r *= k;
  s = r / 2;
  r += k;
  s += max(0, n - r + 1);
  return s;
}
int main() {
  int t, l, r, i, p, m, a, b;
  cin >> t;
  while (t--) {
    cin >> l >> r;
    if (l == r)
      cout << 0 << "\n";
    else {
      p = 1;
      m = r - l;
      for (i = 0; i < 20; i++) {
        a = f(r, p);
        b = f(l - 1, p);
        if (a > b) m = min(m, r - l + 1 - (a - b));
        p *= 2;
      }
      cout << m << "\n";
    }
  }
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int f(int n, int k) {
  int r = n / k, s;
  if (r % 2) r--;
  r *= k;
  s = r / 2;
  r += k;
  s += max(0, n - r + 1);
  return s;
}
int main() {
  int t, l, r, i, p, m, a, b;
  cin >> t;
  while (t--) {
    cin >> l >> r;
    if (l == r)
      cout << 0 << "\n";
    else {
      p = 1;
      m = r - l;
      for (i = 0; i < 20; i++) {
        a = f(r, p);
        b = f(l - 1, p);
        if (a > b) m = min(m, r - l + 1 - (a - b));
        p *= 2;
      }
      cout << m << "\n";
    }
  }
}

```