#include <bits/stdc++.h>
using namespace std;
int mi[30];
int ans, l, r, T;
int get(int x, int t) {
  int ans;
  t = mi[t];
  ans = (x / t / 2) * t;
  x = x % (t * 2);
  if (x >= t) ans += x - t + 1;
  return ans;
}
int main() {
  mi[0] = 1;
  for (int i = 1; i <= 20; i++) mi[i] = mi[i - 1] * 2;
  scanf("%d", &T);
  for (int tt = 1; tt <= T; tt++) {
    ans = 0;
    scanf("%d%d", &l, &r);
    for (int i = 0; i <= 19; i++) {
      ans = max(ans, get(r, i) - get(l - 1, i));
    }
    printf("%d\n", r - l + 1 - ans);
  }
}

### 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;
int mi[30];
int ans, l, r, T;
int get(int x, int t) {
  int ans;
  t = mi[t];
  ans = (x / t / 2) * t;
  x = x % (t * 2);
  if (x >= t) ans += x - t + 1;
  return ans;
}
int main() {
  mi[0] = 1;
  for (int i = 1; i <= 20; i++) mi[i] = mi[i - 1] * 2;
  scanf("%d", &T);
  for (int tt = 1; tt <= T; tt++) {
    ans = 0;
    scanf("%d%d", &l, &r);
    for (int i = 0; i <= 19; i++) {
      ans = max(ans, get(r, i) - get(l - 1, i));
    }
    printf("%d\n", r - l + 1 - ans);
  }
}

```