#include <bits/stdc++.h>
using namespace std;
long long int msbPos(long long n) {
  long long int msb_p = -1;
  while (n) {
    n = n >> 1;
    msb_p++;
  }
  return msb_p;
}
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long res = binpow(a, b / 2);
  if (b % 2)
    return res * res * a;
  else
    return res * res;
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long a = msbPos(l);
  long long b = msbPos(r);
  if (a == b) {
    cout << 0 << endl;
    return;
  } else {
    vector<long long> s(b - a + 1);
    s[0] = (binpow(2, a)) - (l % (binpow(2, a)));
    s[b - a] = (r % (binpow(2, b))) + 1;
    for (long long i = 1; i < b - a; i++) {
      s[i] = binpow(2, a + i);
    }
    vector<long long> v(b + 1, 0);
    for (long long i = 0; i < b - a + 1; i++) {
      for (long long j = 0; j < a + i; j++) {
        long long d = binpow(2, j + 1);
        v[j] += (s[i] / d) * (d / 2);
        if (s[i] % d > d / 2) {
          v[j] += s[i] % d - d / 2;
        }
      }
      v[a + i] += s[i];
    }
    sort(v.rbegin(), v.rend());
    cout << r - l + 1 - v[0] << endl;
  }
}
int main() {
  long long int t;
  cin >> t;
  while (t > 0) {
    solve();
    t--;
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int msbPos(long long n) {
  long long int msb_p = -1;
  while (n) {
    n = n >> 1;
    msb_p++;
  }
  return msb_p;
}
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long res = binpow(a, b / 2);
  if (b % 2)
    return res * res * a;
  else
    return res * res;
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long a = msbPos(l);
  long long b = msbPos(r);
  if (a == b) {
    cout << 0 << endl;
    return;
  } else {
    vector<long long> s(b - a + 1);
    s[0] = (binpow(2, a)) - (l % (binpow(2, a)));
    s[b - a] = (r % (binpow(2, b))) + 1;
    for (long long i = 1; i < b - a; i++) {
      s[i] = binpow(2, a + i);
    }
    vector<long long> v(b + 1, 0);
    for (long long i = 0; i < b - a + 1; i++) {
      for (long long j = 0; j < a + i; j++) {
        long long d = binpow(2, j + 1);
        v[j] += (s[i] / d) * (d / 2);
        if (s[i] % d > d / 2) {
          v[j] += s[i] % d - d / 2;
        }
      }
      v[a + i] += s[i];
    }
    sort(v.rbegin(), v.rend());
    cout << r - l + 1 - v[0] << endl;
  }
}
int main() {
  long long int t;
  cin >> t;
  while (t > 0) {
    solve();
    t--;
  }
}

```