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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int l, r;
    cin >> l >> r;
    int lst[25] = {0};
    for (int i = 0; i < 25; i++) {
      lst[i] = 1 << i;
    }
    int sum[25] = {0};
    for (int i = 0; i < 24; i++) {
      if (l >= lst[i + 1] || r < lst[i]) continue;
      int ll = max(l, lst[i]);
      int rr = min(r, lst[i + 1] - 1);
      sum[i] = rr - ll + 1;
    }
    int maxm = 0;
    for (int i = 0; i < 25; i++) {
      maxm = max(maxm, sum[i]);
    }
    int ans1 = r - l + 1 - maxm;
    int ans2 = 0;
    int num = r - l + 1;
    if (num % 2) {
      if (l % 2 == 0) {
        ans2++;
      }
    }
    ans2 += num / 2;
    printf("%d\n", min(ans1, ans2));
  }
  return 0;
}
