#include <bits/stdc++.h>
using namespace std;
long long count(long long n, long long k) {
  long long res = (n >> (k + 1)) << k;
  if ((n >> k) & 1) res += n & ((1ll << k) - 1);
  return res;
}
void solve() {
  long long l, r;
  cin >> l >> r;
  vector<int> v(21, 0);
  vector<int> u(21, 0);
  vector<int> w(21, 0);
  l--;
  for (int j = 0; j < 21; j++) {
    v[j] = count(r + 1, j);
    u[j] = count(l + 1, j);
    w[j] = v[j] - u[j];
  }
  int n = r - l;
  sort(w.begin(), w.end());
  cout << n - w[20];
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) {
    solve();
    cout << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
long long count(long long n, long long k) {
  long long res = (n >> (k + 1)) << k;
  if ((n >> k) & 1) res += n & ((1ll << k) - 1);
  return res;
}
void solve() {
  long long l, r;
  cin >> l >> r;
  vector<int> v(21, 0);
  vector<int> u(21, 0);
  vector<int> w(21, 0);
  l--;
  for (int j = 0; j < 21; j++) {
    v[j] = count(r + 1, j);
    u[j] = count(l + 1, j);
    w[j] = v[j] - u[j];
  }
  int n = r - l;
  sort(w.begin(), w.end());
  cout << n - w[20];
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) {
    solve();
    cout << "\n";
  }
  return 0;
}

```