#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const long long MAX_N = 1e5 + 1;
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
const long long LINF = 1e18;
void solve() {
  long long l, r;
  cin >> l >> r;
  long long n{};
  long long out = (r - l + 1);
  r++;
  while ((1 << n) <= r) {
    long long a = (r >> (n + 1)) << n;
    if (r & (1 << n)) {
      a += r & ((1 << n) - 1);
    }
    long long b = (l >> (n + 1)) << n;
    if (l & (1 << n)) {
      b += l & ((1 << n) - 1);
    }
    out = min(out, (r - l) - (a - b));
    n++;
  }
  cout << out << '\n';
}
long long main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc;
  cin >> tc;
  for (long long t = 1; t <= tc; 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;
const double PI = acos(-1.0);
const long long MAX_N = 1e5 + 1;
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
const long long LINF = 1e18;
void solve() {
  long long l, r;
  cin >> l >> r;
  long long n{};
  long long out = (r - l + 1);
  r++;
  while ((1 << n) <= r) {
    long long a = (r >> (n + 1)) << n;
    if (r & (1 << n)) {
      a += r & ((1 << n) - 1);
    }
    long long b = (l >> (n + 1)) << n;
    if (l & (1 << n)) {
      b += l & ((1 << n) - 1);
    }
    out = min(out, (r - l) - (a - b));
    n++;
  }
  cout << out << '\n';
}
long long main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc;
  cin >> tc;
  for (long long t = 1; t <= tc; t++) {
    solve();
  }
}

```