#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long max_n = 200005, max_bit = 30;
long long prefix[max_n][max_bit];
void prep() {
  for (long long i = 1; i < max_n; i++) {
    for (long long bit = 0; bit < max_bit; bit++)
      prefix[i][bit] = prefix[i - 1][bit] + ((i >> bit) & 1);
  }
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long max_rem = 0;
  for (long long bit = 0; bit < max_bit; bit++)
    max_rem = max(max_rem, prefix[r][bit] - prefix[l - 1][bit]);
  cout << (r - l + 1) - max_rem << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  prep();
  long long t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long max_n = 200005, max_bit = 30;
long long prefix[max_n][max_bit];
void prep() {
  for (long long i = 1; i < max_n; i++) {
    for (long long bit = 0; bit < max_bit; bit++)
      prefix[i][bit] = prefix[i - 1][bit] + ((i >> bit) & 1);
  }
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long max_rem = 0;
  for (long long bit = 0; bit < max_bit; bit++)
    max_rem = max(max_rem, prefix[r][bit] - prefix[l - 1][bit]);
  cout << (r - l + 1) - max_rem << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  prep();
  long long t;
  cin >> t;
  while (t--) solve();
}

```