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() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  long long int tc;
  cin >> tc;
  while (tc--) {
    long long int l, r;
    cin >> l >> r;
    long long int max2 = log2(r);
    long long int ch = 1;
    vector<long long int> v(max2 + 1, 0);
    v[0] = (r - l) / 2;
    v[0]++;
    for (long long int i = 1; i < max2 + 1; i++) {
      ch = ch << 1;
      long long int s = l, e;
      if (s & ch) {
        while (s <= r && s & ch) {
          v[i]++;
          s++;
        }
        e = r;
        if (e & ch) {
          while (e >= s && e & ch) {
            v[i]++;
            e--;
          }
          e -= ch;
        } else {
          while (e >= s && !(e & ch)) {
            e--;
          }
        }
        if (s < e) {
          v[i] += (e - s) / 2;
        }
      } else {
        while (s <= r && !(s & ch)) {
          s++;
        }
        e = r;
        if (e & ch) {
          while (s <= e && e & ch) {
            v[i]++;
            e--;
          }
        } else {
          while (s <= e && !(e & ch)) {
            e--;
          }
          e -= ch;
          v[i] += ch;
        }
        if (s < e) {
          v[i] += (e - s) / 2;
        }
      }
    }
    cout << r - l + 1 - *max_element(v.begin(), v.end()) << "\n";
  }
}
