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;
vector<long long> dx = {1, -1, 0, 0};
vector<long long> dy = {0, 0, -1, 1};
const long long N = 2e5 + 5;
void solve() {
  long long l, r;
  cin >> l >> r;
  long long a = l;
  long long cnt = (l % 2 == 0) ? 1 : 0;
  for (long long i = l + 1; i < r + 1; i++) {
    if (i % 2 == 0) cnt++;
    a &= i;
  }
  if (a) {
    cout << 0 << '\n';
    return;
  }
  long long one;
  long long cur = 1;
  while (cur <= l) {
    cur *= 2;
  }
  long long r1 = cur - l;
  one = cur;
  cur = 1;
  while (cur <= r) {
    cur *= 2;
  }
  long long r3 = r - cur / 2 + 1;
  long long r2 = cur / 2 - one;
  long long ans = min({r1 + r2, r2 + r3, r1 + r3, cnt});
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
