#include <bits/stdc++.h>
template <class T>
int gmax(T &a, T b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
int gmin(T &a, T b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
using namespace std;
const int INF = 0x3f3f3f3f;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int n = r - l + 1;
    int result = INF;
    for (int i = 2; i < 1e6; i <<= 1) {
      int left_num = l / i * i / 2 + max(0, l % i - i / 2);
      int right_num = (r + 1) / i * i / 2 + max(0, (r + 1) % i - i / 2);
      result = min(result, n - right_num + left_num);
      if (right_num == 0) break;
    }
    printf("%d\n", result);
  }
  return 0;
}

### Prompt

Please create a solution in cpp to 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>
template <class T>
int gmax(T &a, T b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
int gmin(T &a, T b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
using namespace std;
const int INF = 0x3f3f3f3f;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int n = r - l + 1;
    int result = INF;
    for (int i = 2; i < 1e6; i <<= 1) {
      int left_num = l / i * i / 2 + max(0, l % i - i / 2);
      int right_num = (r + 1) / i * i / 2 + max(0, (r + 1) % i - i / 2);
      result = min(result, n - right_num + left_num);
      if (right_num == 0) break;
    }
    printf("%d\n", result);
  }
  return 0;
}

```