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;
void c_p_c() {
  clock_t begin = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void c_p_c2() {
  clock_t begin = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long log(long long n) {
  long long ctr = 0;
  while (n) {
    ctr += 1;
    n >>= 1;
  }
  return ctr;
}
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long len = r - l + 1;
  long long start = l;
  long long nxt = binpow(2, log2(l) + 1);
  long long ans = INT_MAX;
  while (true) {
    if (nxt < r) {
      long long currLen = nxt - start;
      ans = min(ans, len - currLen);
    } else if (nxt > r) {
      long long currLen = r - start + 1;
      ans = min(ans, len - currLen);
      break;
    }
    start = nxt;
    nxt *= 2;
  }
  ans = min(ans, len / 2);
  cout << ans << '\n';
}
int32_t main() {
  c_p_c();
  long long t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    solve();
  }
  c_p_c2();
}
