#include <bits/stdc++.h>
using namespace std;
int a[20][200005];
void solve() {
  int l, r;
  cin >> l >> r;
  int maxi = 0;
  for (int i = 0; i < 20; i++) maxi = max(maxi, a[i][r] - a[i][l - 1]);
  cout << (r - l + 1) - maxi << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 1; i <= 200000; i++) {
    for (int j = 0; j < 20; j++) {
      a[j][i] += a[j][i - 1];
      if ((i >> j) & 1) a[j][i]++;
    }
  }
  while (t--) solve();
  return 0;
}

### Prompt

Generate 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 a[20][200005];
void solve() {
  int l, r;
  cin >> l >> r;
  int maxi = 0;
  for (int i = 0; i < 20; i++) maxi = max(maxi, a[i][r] - a[i][l - 1]);
  cout << (r - l + 1) - maxi << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 1; i <= 200000; i++) {
    for (int j = 0; j < 20; j++) {
      a[j][i] += a[j][i - 1];
      if ((i >> j) & 1) a[j][i]++;
    }
  }
  while (t--) solve();
  return 0;
}

```