#include <bits/stdc++.h>
using namespace std;
int a[200001][32];
int on[200001][32];
void solve() {
  int le, r;
  cin >> le >> r;
  int oc[32];
  for (int i = 0; i < 32; i++) {
    oc[i] = on[r][i] - on[le - 1][i];
  }
  sort(oc, oc + 32, greater<int>());
  cout << r - le + 1 - oc[0] << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < 200001; i++) {
    int c = 0;
    for (int j = 31; j >= 0; j--) {
      if (((1 << j) & i) == 0) {
        a[i][c] = 0;
      } else {
        a[i][c] = 1;
      }
      c++;
    }
  }
  for (int j = 0; j < 32; j++) {
    int c = 0;
    for (int i = 0; i < 200001; i++) {
      if (a[i][j] == 1) {
        c++;
      }
      on[i][j] = c;
    }
  }
  while (t--) {
    solve();
  }
  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>
using namespace std;
int a[200001][32];
int on[200001][32];
void solve() {
  int le, r;
  cin >> le >> r;
  int oc[32];
  for (int i = 0; i < 32; i++) {
    oc[i] = on[r][i] - on[le - 1][i];
  }
  sort(oc, oc + 32, greater<int>());
  cout << r - le + 1 - oc[0] << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < 200001; i++) {
    int c = 0;
    for (int j = 31; j >= 0; j--) {
      if (((1 << j) & i) == 0) {
        a[i][c] = 0;
      } else {
        a[i][c] = 1;
      }
      c++;
    }
  }
  for (int j = 0; j < 32; j++) {
    int c = 0;
    for (int i = 0; i < 200001; i++) {
      if (a[i][j] == 1) {
        c++;
      }
      on[i][j] = c;
    }
  }
  while (t--) {
    solve();
  }
  return 0;
}

```