#include <bits/stdc++.h>
using namespace std;
long long cnt(long long n, long long x) {
  n++;
  long long l = n / x;
  long long m = (n - (l * x));
  long long ans = (l / 2) * x;
  if (l % 2 == 1) ans += m;
  return ans;
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long tot = r - l + 1;
  long long ans = tot - 1;
  long long p2 = 1;
  while (p2 <= r) {
    long long tmp = cnt(r, p2) - cnt(l - 1, p2);
    tmp = tot - tmp;
    ans = min(ans, tmp);
    p2 *= 2;
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### 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;
long long cnt(long long n, long long x) {
  n++;
  long long l = n / x;
  long long m = (n - (l * x));
  long long ans = (l / 2) * x;
  if (l % 2 == 1) ans += m;
  return ans;
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long tot = r - l + 1;
  long long ans = tot - 1;
  long long p2 = 1;
  while (p2 <= r) {
    long long tmp = cnt(r, p2) - cnt(l - 1, p2);
    tmp = tot - tmp;
    ans = min(ans, tmp);
    p2 *= 2;
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```