#include <bits/stdc++.h>
using namespace std;
vector<int> two;
void make_two(vector<int>& two) {
  int t = 1;
  for (int i = 0; i < 19; i++) {
    two.emplace_back(t);
    t *= 2;
  }
  return;
}
int zero_cnt(int bit, int num) {
  int cycle = two[bit + 1];
  int q = num / cycle;
  int r = num % cycle;
  int ret = two[bit] * q;
  if (r < two[bit]) {
    ret += r + 1;
    return ret;
  } else {
    ret += two[bit];
    return ret;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  make_two(two);
  while (t--) {
    int l, r;
    cin >> l >> r;
    int minval = INT_MAX;
    for (int bit = 0; bit < 18; bit++) {
      int tmp = zero_cnt(bit, r) - zero_cnt(bit, l - 1);
      if (tmp < minval) minval = tmp;
    }
    cout << minval << "\n";
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
vector<int> two;
void make_two(vector<int>& two) {
  int t = 1;
  for (int i = 0; i < 19; i++) {
    two.emplace_back(t);
    t *= 2;
  }
  return;
}
int zero_cnt(int bit, int num) {
  int cycle = two[bit + 1];
  int q = num / cycle;
  int r = num % cycle;
  int ret = two[bit] * q;
  if (r < two[bit]) {
    ret += r + 1;
    return ret;
  } else {
    ret += two[bit];
    return ret;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  make_two(two);
  while (t--) {
    int l, r;
    cin >> l >> r;
    int minval = INT_MAX;
    for (int bit = 0; bit < 18; bit++) {
      int tmp = zero_cnt(bit, r) - zero_cnt(bit, l - 1);
      if (tmp < minval) minval = tmp;
    }
    cout << minval << "\n";
  }
  return 0;
}

```