#include <bits/stdc++.h>
const int mod = 1e9 + 7;
const long long maxx = 1e18;
using namespace std;
const int N = 2e5 + 5;
vector<vector<int>> res(N, vector<int>(20, 0));
void preCompute() {
  for (int i = 0; i < 20; i++) {
    int mask = 1 << i;
    int cnt = 0;
    for (int j = 1; j < N; j++) {
      if ((mask & j))
        res[j][i] = res[j - 1][i] + 1;
      else
        res[j][i] = res[j - 1][i];
    }
  }
}
void solve() {
  int l, r;
  cin >> l >> r;
  int mx = INT_MIN;
  for (int i = 0; i < 20; i++) mx = max(res[r][i] - res[l - 1][i], mx);
  cout << r - l + 1 - mx << "\n";
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  int t;
  preCompute();
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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>
const int mod = 1e9 + 7;
const long long maxx = 1e18;
using namespace std;
const int N = 2e5 + 5;
vector<vector<int>> res(N, vector<int>(20, 0));
void preCompute() {
  for (int i = 0; i < 20; i++) {
    int mask = 1 << i;
    int cnt = 0;
    for (int j = 1; j < N; j++) {
      if ((mask & j))
        res[j][i] = res[j - 1][i] + 1;
      else
        res[j][i] = res[j - 1][i];
    }
  }
}
void solve() {
  int l, r;
  cin >> l >> r;
  int mx = INT_MIN;
  for (int i = 0; i < 20; i++) mx = max(res[r][i] - res[l - 1][i], mx);
  cout << r - l + 1 - mx << "\n";
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  int t;
  preCompute();
  cin >> t;
  while (t--) solve();
  return 0;
}

```