#include <bits/stdc++.h>
using namespace std;
template <typename... Args>
void logger(string vars, Args &&...values) {
  cout << vars << " = ";
  string delim = "";
  (..., (cout << delim << values, delim = ", "));
}
template <typename T>
void print(vector<T> &collection, bool newline = true) {
  for (auto &item : collection) {
    cout << item << " ";
  }
  if (newline) cout << "\n";
}
template <typename T>
void print(T _data, bool newline = true) {
  cout << _data;
  if (newline) cout << "\n";
}
long long f;
long long n, w, k, x, m, t, p;
long long cnt = 0;
long long getSetBitsFromOneToN(long long r, long long l) {
  long long two = 2, ans = INT_MAX;
  long long n1 = r, n2 = l;
  l--;
  long long n = r;
  while (n) {
    long long curr = 0;
    curr += (r / two) * (two >> 1);
    if ((r & (two - 1)) > (two >> 1) - 1)
      curr += (r & (two - 1)) - (two >> 1) + 1;
    n >>= 1;
    if (true) {
      curr -= (l / two) * (two >> 1);
      if ((l & (two - 1)) > (two >> 1) - 1)
        curr -= (l & (two - 1)) - (two >> 1) + 1;
    }
    two <<= 1;
    ans = min(ans, n1 - n2 + 1 - curr);
  }
  return ans;
}
long long helper(long long l, long long r) {
  long long ans = INT_MAX;
  for (long long bit = 31; bit >= 0; bit--) {
    long long curr = 0;
    for (long long num = l; num <= r; num++) {
      if (num & (1 << bit)) curr++;
    }
    ans = min(ans, r - l + 1 - curr);
  }
  return ans;
}
void Solve() {
  long long l, r;
  cin >> l >> r;
  cout << getSetBitsFromOneToN(r, l);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long test;
  cin >> test;
  for (long long _ = 0; _ < test; _++) {
    Solve();
    cout << "\n";
  }
}

### Prompt

Create a solution in Cpp for the following problem:

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;
template <typename... Args>
void logger(string vars, Args &&...values) {
  cout << vars << " = ";
  string delim = "";
  (..., (cout << delim << values, delim = ", "));
}
template <typename T>
void print(vector<T> &collection, bool newline = true) {
  for (auto &item : collection) {
    cout << item << " ";
  }
  if (newline) cout << "\n";
}
template <typename T>
void print(T _data, bool newline = true) {
  cout << _data;
  if (newline) cout << "\n";
}
long long f;
long long n, w, k, x, m, t, p;
long long cnt = 0;
long long getSetBitsFromOneToN(long long r, long long l) {
  long long two = 2, ans = INT_MAX;
  long long n1 = r, n2 = l;
  l--;
  long long n = r;
  while (n) {
    long long curr = 0;
    curr += (r / two) * (two >> 1);
    if ((r & (two - 1)) > (two >> 1) - 1)
      curr += (r & (two - 1)) - (two >> 1) + 1;
    n >>= 1;
    if (true) {
      curr -= (l / two) * (two >> 1);
      if ((l & (two - 1)) > (two >> 1) - 1)
        curr -= (l & (two - 1)) - (two >> 1) + 1;
    }
    two <<= 1;
    ans = min(ans, n1 - n2 + 1 - curr);
  }
  return ans;
}
long long helper(long long l, long long r) {
  long long ans = INT_MAX;
  for (long long bit = 31; bit >= 0; bit--) {
    long long curr = 0;
    for (long long num = l; num <= r; num++) {
      if (num & (1 << bit)) curr++;
    }
    ans = min(ans, r - l + 1 - curr);
  }
  return ans;
}
void Solve() {
  long long l, r;
  cin >> l >> r;
  cout << getSetBitsFromOneToN(r, l);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long test;
  cin >> test;
  for (long long _ = 0; _ < test; _++) {
    Solve();
    cout << "\n";
  }
}

```