#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = ((res % p) * (x % p)) % p;
    y = y >> 1;
    x = ((x % p) * (x % p)) % p;
  }
  return res % p;
}
long long int previouss[200005][31];
void precalculate() {
  long long int N = 2e5 + 5;
  for (long long int i = 1; i < N; i++) {
    for (int j = 0; j <= 30; j++) {
      if (i & (1 << j)) previouss[i][j] = 1;
      previouss[i][j] += previouss[i - 1][j];
    }
  }
}
void solve() {
  long long int n, m;
  cin >> n >> m;
  long long int total = m - n + 1;
  long long int ans = 1e10;
  for (int j = 0; j <= 30; j++) {
    long long int curr = (previouss[m][j] - previouss[n - 1][j]);
    ans = min(ans, total - curr);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  precalculate();
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = ((res % p) * (x % p)) % p;
    y = y >> 1;
    x = ((x % p) * (x % p)) % p;
  }
  return res % p;
}
long long int previouss[200005][31];
void precalculate() {
  long long int N = 2e5 + 5;
  for (long long int i = 1; i < N; i++) {
    for (int j = 0; j <= 30; j++) {
      if (i & (1 << j)) previouss[i][j] = 1;
      previouss[i][j] += previouss[i - 1][j];
    }
  }
}
void solve() {
  long long int n, m;
  cin >> n >> m;
  long long int total = m - n + 1;
  long long int ans = 1e10;
  for (int j = 0; j <= 30; j++) {
    long long int curr = (previouss[m][j] - previouss[n - 1][j]);
    ans = min(ans, total - curr);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  precalculate();
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```