#include <bits/stdc++.h>
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const long long int int_max = 1e18;
using namespace std;
void file_i_o() {
  ios::sync_with_stdio(false), cin.tie(0);
  cout.tie(0);
}
vector<vector<long long int>> bset(2e5 + 2, vector<long long int>(31, 0));
void precompute() {
  long long int n = 2e5 + 1;
  for (long long int i = 0; i <= n; i++) {
    for (long long int j = 0; j <= 30; j++) {
      if ((1 << j) & i) {
        if (i != 0)
          bset[i][j] = bset[i - 1][j] + 1;
        else
          bset[i][j] = 1;
      } else {
        if (i != 0)
          bset[i][j] += bset[i - 1][j];
        else
          bset[i][j] = 0;
      }
    }
  }
}
void solve() {
  long long int l, r;
  cin >> l >> r;
  long long int ans = int_max;
  for (long long int i = 0; i <= 30; i++) {
    if (l - 1 >= 0)
      ans = min(ans, (r - l + 1) - (bset[r][i] - bset[l - 1][i]));
    else
      ans = min(ans, (r - l + 1) - bset[r][i]);
  }
  cout << ans << endl;
}
int main(int argc, char const *argv[]) {
  file_i_o();
  long long int t;
  t = 1;
  cin >> t;
  precompute();
  while (t--) solve();
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const long long int int_max = 1e18;
using namespace std;
void file_i_o() {
  ios::sync_with_stdio(false), cin.tie(0);
  cout.tie(0);
}
vector<vector<long long int>> bset(2e5 + 2, vector<long long int>(31, 0));
void precompute() {
  long long int n = 2e5 + 1;
  for (long long int i = 0; i <= n; i++) {
    for (long long int j = 0; j <= 30; j++) {
      if ((1 << j) & i) {
        if (i != 0)
          bset[i][j] = bset[i - 1][j] + 1;
        else
          bset[i][j] = 1;
      } else {
        if (i != 0)
          bset[i][j] += bset[i - 1][j];
        else
          bset[i][j] = 0;
      }
    }
  }
}
void solve() {
  long long int l, r;
  cin >> l >> r;
  long long int ans = int_max;
  for (long long int i = 0; i <= 30; i++) {
    if (l - 1 >= 0)
      ans = min(ans, (r - l + 1) - (bset[r][i] - bset[l - 1][i]));
    else
      ans = min(ans, (r - l + 1) - bset[r][i]);
  }
  cout << ans << endl;
}
int main(int argc, char const *argv[]) {
  file_i_o();
  long long int t;
  t = 1;
  cin >> t;
  precompute();
  while (t--) solve();
  return 0;
}

```