#include <bits/stdc++.h>
using namespace std;
int min(int a, int b) {
  if (a < b)
    return a;
  else
    return b;
}
void printArray(int arr[], int size) {
  for (int i = 0; i < size; i++) cout << arr[i] << " ";
  cout << endl;
}
int minimum(int arr[], int n) {
  int min = arr[0];
  for (int i = 1; i < n; i++)
    if (arr[i] < min) min = arr[i];
  return min;
}
int maximum(int arr[], int n) {
  int max = arr[0];
  for (int i = 1; i < n; i++)
    if (arr[i] > max) max = arr[i];
  return max;
}
int maxSubarraySum(int arr[], int n) {
  int ms = INT_MIN;
  int me = 0;
  int start = 0;
  int end = 0;
  int s = 0;
  for (int i = 0; i < n; i++) {
    me += arr[i];
    if (me > ms) {
      ms = me;
      start = s;
      end = i;
    }
    if (me < 0) {
      me = 0;
      s = i + 1;
    }
  }
  return ms;
}
bool sortcol(const vector<int>& v1, const vector<int>& v2) {
  return v1[0] < v2[0];
}
vector<vector<int>> dp(200001, vector<int>(32));
void start() {
  for (int i = 1; i < 200001; i++) {
    dp[i] = dp[i - 1];
    string x = std::bitset<32>(i).to_string();
    reverse(x.begin(), x.end());
    for (int j = 0; j < x.size(); j++) {
      if (x[j] == '1') dp[i][j]++;
    }
  }
}
int main() {
  start();
  int t = 1;
  cin >> t;
  while (t--) {
    int l, r;
    cin >> l >> r;
    int n = r - l + 1;
    int tog = 2;
    for (; tog < l; tog *= 2)
      ;
    int frog = 1;
    for (; frog <= r; frog *= 2)
      ;
    int first = tog - l;
    int last = r - frog / 2 + 1;
    int var = frog / 4;
    int miniMUM = INT_MAX;
    for (int i = 0; i < 32; i++) {
      int VOF = dp[r][i] - dp[l - 1][i];
      int zero = (r - l + 1) - VOF;
      miniMUM = min(miniMUM, zero);
    }
    cout << miniMUM << endl;
  }
  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 min(int a, int b) {
  if (a < b)
    return a;
  else
    return b;
}
void printArray(int arr[], int size) {
  for (int i = 0; i < size; i++) cout << arr[i] << " ";
  cout << endl;
}
int minimum(int arr[], int n) {
  int min = arr[0];
  for (int i = 1; i < n; i++)
    if (arr[i] < min) min = arr[i];
  return min;
}
int maximum(int arr[], int n) {
  int max = arr[0];
  for (int i = 1; i < n; i++)
    if (arr[i] > max) max = arr[i];
  return max;
}
int maxSubarraySum(int arr[], int n) {
  int ms = INT_MIN;
  int me = 0;
  int start = 0;
  int end = 0;
  int s = 0;
  for (int i = 0; i < n; i++) {
    me += arr[i];
    if (me > ms) {
      ms = me;
      start = s;
      end = i;
    }
    if (me < 0) {
      me = 0;
      s = i + 1;
    }
  }
  return ms;
}
bool sortcol(const vector<int>& v1, const vector<int>& v2) {
  return v1[0] < v2[0];
}
vector<vector<int>> dp(200001, vector<int>(32));
void start() {
  for (int i = 1; i < 200001; i++) {
    dp[i] = dp[i - 1];
    string x = std::bitset<32>(i).to_string();
    reverse(x.begin(), x.end());
    for (int j = 0; j < x.size(); j++) {
      if (x[j] == '1') dp[i][j]++;
    }
  }
}
int main() {
  start();
  int t = 1;
  cin >> t;
  while (t--) {
    int l, r;
    cin >> l >> r;
    int n = r - l + 1;
    int tog = 2;
    for (; tog < l; tog *= 2)
      ;
    int frog = 1;
    for (; frog <= r; frog *= 2)
      ;
    int first = tog - l;
    int last = r - frog / 2 + 1;
    int var = frog / 4;
    int miniMUM = INT_MAX;
    for (int i = 0; i < 32; i++) {
      int VOF = dp[r][i] - dp[l - 1][i];
      int zero = (r - l + 1) - VOF;
      miniMUM = min(miniMUM, zero);
    }
    cout << miniMUM << endl;
  }
  return 0;
}

```