#include <bits/stdc++.h>
using namespace std;
const int t_max = 1e9 + 7;
const long long MAXX = long long(1e18);
const int INF = int(-1e9);
const double PI = 3.1415926535;
void read() {}
template <typename T1>
void read(T1 v[], int n) {
  for (int i = 0; i < n; i++) cin >> v[i];
}
template <typename T2>
void read(vector<T2>& v, int n) {
  for (int i = 0; i < v.size(); i++) cin >> v[i];
}
template <typename First, typename... Rest>
void read(First& first, Rest&... rest) {
  cin >> first;
  read(rest...);
}
template <typename T3>
void print(vector<T3>& v, int n) {
  for (int i = 0; i < v.size(); i++) cout << v[i] << '\n';
}
template <typename T4>
void print(T4 v[], int n) {
  for (int i = 0; i < n; i++) cout << v[i] << '\n';
}
long long binpow(long long a, long long n, long long k) {
  long long res = 1;
  while (n)
    if (n & 1) {
      res *= a;
      res %= k;
      n--;
    } else {
      a *= a;
      a %= k;
      n >>= 1;
    }
  return res;
}
const int M = 200001, L = 23;
int arr[M][L];
void solve() {
  int l, r;
  cin >> l >> r;
  int maxx = 0;
  for (int i = 0; i < L; i++) maxx = max(maxx, arr[r][i] - arr[l - 1][i]);
  cout << r - l + 1 - maxx << '\n';
}
int main() {
  for (int i = 0; i < L; i++) arr[0][i] = 0;
  for (int i = 1; i < M; i++) {
    for (int j = 0; j < L; j++) arr[i][j] = arr[i - 1][j];
    int k = i;
    int pos = 0;
    while (k != 0) {
      arr[i][pos] += k % 2;
      k /= 2;
      pos++;
    }
  }
  cin.tie(0);
  cout.tie(0);
  std::ios_base::sync_with_stdio(0);
  int t = 1;
  cin >> t;
  for (int i = 0; i < t; i++) {
    solve();
  }
}

### Prompt

Your task is to create a CPP solution 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;
const int t_max = 1e9 + 7;
const long long MAXX = long long(1e18);
const int INF = int(-1e9);
const double PI = 3.1415926535;
void read() {}
template <typename T1>
void read(T1 v[], int n) {
  for (int i = 0; i < n; i++) cin >> v[i];
}
template <typename T2>
void read(vector<T2>& v, int n) {
  for (int i = 0; i < v.size(); i++) cin >> v[i];
}
template <typename First, typename... Rest>
void read(First& first, Rest&... rest) {
  cin >> first;
  read(rest...);
}
template <typename T3>
void print(vector<T3>& v, int n) {
  for (int i = 0; i < v.size(); i++) cout << v[i] << '\n';
}
template <typename T4>
void print(T4 v[], int n) {
  for (int i = 0; i < n; i++) cout << v[i] << '\n';
}
long long binpow(long long a, long long n, long long k) {
  long long res = 1;
  while (n)
    if (n & 1) {
      res *= a;
      res %= k;
      n--;
    } else {
      a *= a;
      a %= k;
      n >>= 1;
    }
  return res;
}
const int M = 200001, L = 23;
int arr[M][L];
void solve() {
  int l, r;
  cin >> l >> r;
  int maxx = 0;
  for (int i = 0; i < L; i++) maxx = max(maxx, arr[r][i] - arr[l - 1][i]);
  cout << r - l + 1 - maxx << '\n';
}
int main() {
  for (int i = 0; i < L; i++) arr[0][i] = 0;
  for (int i = 1; i < M; i++) {
    for (int j = 0; j < L; j++) arr[i][j] = arr[i - 1][j];
    int k = i;
    int pos = 0;
    while (k != 0) {
      arr[i][pos] += k % 2;
      k /= 2;
      pos++;
    }
  }
  cin.tie(0);
  cout.tie(0);
  std::ios_base::sync_with_stdio(0);
  int t = 1;
  cin >> t;
  for (int i = 0; i < t; i++) {
    solve();
  }
}

```