#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k, t, m, p;
  cin >> p;
  for (k = 0; k < p; k++) {
    cin >> m >> n;
    int x, y;
    int z = (n - m + 1);
    n = n + 1;
    int minT = z - (n / 2) + (m / 2);
    int pow2 = 2;
    while (pow2 <= m || pow2 <= n) {
      int c1 = (m / pow2);
      int c2 = (n / pow2);
      x = (c1 / 2) * pow2;
      y = (c2 / 2) * pow2;
      x += (c1 % 2 == 0) ? 0 : (m % pow2);
      y += (c2 % 2 == 0) ? 0 : (n % pow2);
      minT = min(minT, z - y + x);
      pow2 = pow2 * 2;
    }
    cout << minT << endl;
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 main() {
  int n, k, t, m, p;
  cin >> p;
  for (k = 0; k < p; k++) {
    cin >> m >> n;
    int x, y;
    int z = (n - m + 1);
    n = n + 1;
    int minT = z - (n / 2) + (m / 2);
    int pow2 = 2;
    while (pow2 <= m || pow2 <= n) {
      int c1 = (m / pow2);
      int c2 = (n / pow2);
      x = (c1 / 2) * pow2;
      y = (c2 / 2) * pow2;
      x += (c1 % 2 == 0) ? 0 : (m % pow2);
      y += (c2 % 2 == 0) ? 0 : (n % pow2);
      minT = min(minT, z - y + x);
      pow2 = pow2 * 2;
    }
    cout << minT << endl;
  }
}

```