#include <bits/stdc++.h>
using namespace std;
void fast_io() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
mt19937 rng(std::chrono::steady_clock::now().time_since_epoch().count());
int main() {
  fast_io();
  int test;
  cin >> test;
  vector<vector<int> > pref(3e5, vector<int>(20));
  for (int i = 1; i <= 3e5; i++) {
    pref[i] = pref[i - 1];
    for (int j = 0; j < 20; j++) pref[i][j] += (i >> j) % 2;
  }
  for (int tc = 0; tc < test; tc++) {
    int l, r;
    cin >> l >> r;
    vector<int> bits = pref[r];
    for (int j = 0; j < 20; j++) bits[j] -= pref[l - 1][j];
    int best = 0;
    for (int j = 0; j < 20; j++) best = max(best, bits[j]);
    cout << r - l + 1 - best << "\n";
  }
}

### Prompt

Please provide a Cpp coded solution 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;
void fast_io() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
mt19937 rng(std::chrono::steady_clock::now().time_since_epoch().count());
int main() {
  fast_io();
  int test;
  cin >> test;
  vector<vector<int> > pref(3e5, vector<int>(20));
  for (int i = 1; i <= 3e5; i++) {
    pref[i] = pref[i - 1];
    for (int j = 0; j < 20; j++) pref[i][j] += (i >> j) % 2;
  }
  for (int tc = 0; tc < test; tc++) {
    int l, r;
    cin >> l >> r;
    vector<int> bits = pref[r];
    for (int j = 0; j < 20; j++) bits[j] -= pref[l - 1][j];
    int best = 0;
    for (int j = 0; j < 20; j++) best = max(best, bits[j]);
    cout << r - l + 1 - best << "\n";
  }
}

```