#include <bits/stdc++.h>
using namespace std;
const char nl = '\n';
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
int N = 2e5 + 2;
vector<vector<int>> pre(N, vector<int>(21, 0));
void precalc() {
  for (int i = 1; i < N; ++i) {
    for (int j = 0; j < 21; ++j) {
      if ((i & (1 << j))) pre[i][j] = 1;
      pre[i][j] += pre[i - 1][j];
    }
  }
}
void solve() {
  int l, r;
  cin >> l >> r;
  int ans = 0;
  for (int i = 0; i < 21; ++i) {
    ans = max(ans, pre[r][i] - pre[l - 1][i]);
  }
  cout << r - l + 1 - ans << nl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  precalc();
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### 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;
const char nl = '\n';
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
int N = 2e5 + 2;
vector<vector<int>> pre(N, vector<int>(21, 0));
void precalc() {
  for (int i = 1; i < N; ++i) {
    for (int j = 0; j < 21; ++j) {
      if ((i & (1 << j))) pre[i][j] = 1;
      pre[i][j] += pre[i - 1][j];
    }
  }
}
void solve() {
  int l, r;
  cin >> l >> r;
  int ans = 0;
  for (int i = 0; i < 21; ++i) {
    ans = max(ans, pre[r][i] - pre[l - 1][i]);
  }
  cout << r - l + 1 - ans << nl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  precalc();
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```