#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 1;
const int rN = 1e5 + 1;
constexpr long double PI = 3.141592653589793238462;
template <typename T1, typename T2>
istream &operator>>(istream &istream, pair<T1, T2> &p) {
  return (istream >> p.first >> p.second);
}
template <typename T>
istream &operator>>(istream &istream, vector<T> &v) {
  for (auto &it : v) cin >> it;
  return istream;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &ostream, const pair<T1, T2> &p) {
  return (ostream << p.first << " " << p.second);
}
template <typename T>
ostream &operator<<(ostream &ostream, const vector<T> &c) {
  for (auto &it : c) cout << it << " ";
  return ostream;
}
bool prime[N];
void process() {
  for (int i = 2; i < N; i++) prime[i] = true;
  prime[0] = prime[1] = false;
  for (int i = 2; i * i <= N; i++) {
    if (prime[i] == true) {
      for (int j = i * i; j < N; j += i) {
        prime[j] = false;
      }
    }
  }
}
long long power(long long a, long long n, long long num) {
  long long res = a;
  long long x = 1;
  while (n) {
    if (n % 2 == 1) x = (x * res) % num;
    res = (res * res) % num;
    n /= 2;
  }
  return x % num;
}
long long inverse(long long n, long long num) { return power(n, num - 2, num); }
int gcd(int a, int b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
int lcm(int a, int b) { return (a / gcd(a, b)) * b; }
int dp[31][200001];
void solve() {
  int l, r;
  cin >> l >> r;
  int ans = INT_MAX;
  for (int i = 0; i < 31; i++) {
    ans = min(ans, dp[i][r] - dp[i][l - 1]);
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  memset((dp), 0, sizeof((dp)));
  int temp;
  int idx;
  for (int i = 1; i < 200001; i++) {
    for (int j = 0; j < 31; j++) {
      dp[j][i] += dp[j][i - 1] + (!(i & (1 << j)));
    }
  }
  int tt;
  cin >> tt;
  while (tt--) solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 N = 1e6 + 1;
const int rN = 1e5 + 1;
constexpr long double PI = 3.141592653589793238462;
template <typename T1, typename T2>
istream &operator>>(istream &istream, pair<T1, T2> &p) {
  return (istream >> p.first >> p.second);
}
template <typename T>
istream &operator>>(istream &istream, vector<T> &v) {
  for (auto &it : v) cin >> it;
  return istream;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &ostream, const pair<T1, T2> &p) {
  return (ostream << p.first << " " << p.second);
}
template <typename T>
ostream &operator<<(ostream &ostream, const vector<T> &c) {
  for (auto &it : c) cout << it << " ";
  return ostream;
}
bool prime[N];
void process() {
  for (int i = 2; i < N; i++) prime[i] = true;
  prime[0] = prime[1] = false;
  for (int i = 2; i * i <= N; i++) {
    if (prime[i] == true) {
      for (int j = i * i; j < N; j += i) {
        prime[j] = false;
      }
    }
  }
}
long long power(long long a, long long n, long long num) {
  long long res = a;
  long long x = 1;
  while (n) {
    if (n % 2 == 1) x = (x * res) % num;
    res = (res * res) % num;
    n /= 2;
  }
  return x % num;
}
long long inverse(long long n, long long num) { return power(n, num - 2, num); }
int gcd(int a, int b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
int lcm(int a, int b) { return (a / gcd(a, b)) * b; }
int dp[31][200001];
void solve() {
  int l, r;
  cin >> l >> r;
  int ans = INT_MAX;
  for (int i = 0; i < 31; i++) {
    ans = min(ans, dp[i][r] - dp[i][l - 1]);
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  memset((dp), 0, sizeof((dp)));
  int temp;
  int idx;
  for (int i = 1; i < 200001; i++) {
    for (int j = 0; j < 31; j++) {
      dp[j][i] += dp[j][i - 1] + (!(i & (1 << j)));
    }
  }
  int tt;
  cin >> tt;
  while (tt--) solve();
  return 0;
}

```