#include <bits/stdc++.h>
using namespace std;
int main() {
  int pp;
  cin >> pp;
  for (int jiaran = 1; jiaran <= pp; jiaran++) {
    int left, right, t = 1;
    cin >> left >> right;
    int p1 = 1, p2 = 2, temp = 0, p;
    int l, r, k1, k2, k;
    while (p2 <= right * 2) {
      p = 0;
      l = left / p1 + (left % p1 > 0);
      r = right / p1;
      if (r >= l) {
        p = (r - l);
        p /= 2;
        if (l % 2 > 0 && r % 2 == 0) p++;
        p *= p1;
        if (l % 2 == 0) p += l * p1 - left;
        if (r % 2 != 0) p += right - r * p1 + 1;
      } else {
        if (r % 2 > 0) p = right - left + 1;
      }
      if (p > temp) {
        temp = p;
      }
      p1 *= 2;
      p2 *= 2;
    }
    cout << right - left + 1 - temp << endl << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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 main() {
  int pp;
  cin >> pp;
  for (int jiaran = 1; jiaran <= pp; jiaran++) {
    int left, right, t = 1;
    cin >> left >> right;
    int p1 = 1, p2 = 2, temp = 0, p;
    int l, r, k1, k2, k;
    while (p2 <= right * 2) {
      p = 0;
      l = left / p1 + (left % p1 > 0);
      r = right / p1;
      if (r >= l) {
        p = (r - l);
        p /= 2;
        if (l % 2 > 0 && r % 2 == 0) p++;
        p *= p1;
        if (l % 2 == 0) p += l * p1 - left;
        if (r % 2 != 0) p += right - r * p1 + 1;
      } else {
        if (r % 2 > 0) p = right - left + 1;
      }
      if (p > temp) {
        temp = p;
      }
      p1 *= 2;
      p2 *= 2;
    }
    cout << right - left + 1 - temp << endl << endl;
  }
  return 0;
}

```