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;
const int mod = (int)(1e9 + 7);
const int N = 200010;
const int inf = (int)(1e9);
void solve() {
  long long l, r;
  cin >> l >> r;
  long long i, fact = 1;
  for (i = 0; i < 30; i++) {
    if (fact > r) {
      break;
    }
    fact = fact * 2;
  }
  long long pow = i;
  pow--;
  fact = fact / 2;
  long long ans = r - fact + 1;
  ans = r - l + 1 - ans;
  if (fact / 2 >= l) {
    ans = min(ans, r - l + 1 - fact / 2);
  }
  if ((r - l + 1) % 2) {
    if (r % 2) {
      ans = min(ans, (r - l + 1) / 2);
    } else {
      ans = min(ans, (r - l + 1) / 2 + 1);
    }
  } else {
    ans = min(ans, (r - l + 1) / 2 + 1);
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tt = 1;
  cin >> tt;
  while (tt--) solve();
  return 0;
}
