#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
const int N = (int)1e5 + 1;
void TimeExecution() {
  cout << "\n\n===>> Time Execution: " << clock() / (double)1000 << " sec(s).";
}
inline long long fpow(long long n, long long k) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n;
    n = n * n;
  }
  return r;
}
long long sqr(long long a) { return a * a; }
long long msqr(long long a) { return ((a % MOD) * (a % MOD)) % MOD; }
long long msub(long long a, long long b) { return ((a - b) % MOD + MOD) % MOD; }
long long madd(long long a, long long b) { return (a + b) % MOD; }
long long mmul(long long a, long long b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
int cal(int x, int s) {
  int cnt = 0;
  int len = (1 << s);
  return (x / len + 1) * (len / 2) -
         max(0, ((x / len) * len + (len / 2) - 1 - x));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int l, r;
    cin >> l >> r;
    int ans = INF;
    int len = floor(log2(r)) + 1;
    while (len) {
      int v1 = cal(r, len);
      int v2 = cal(l - 1, len);
      ans = min(ans, cal(r, len) - cal(l - 1, len));
      len--;
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please formulate 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 MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
const int N = (int)1e5 + 1;
void TimeExecution() {
  cout << "\n\n===>> Time Execution: " << clock() / (double)1000 << " sec(s).";
}
inline long long fpow(long long n, long long k) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n;
    n = n * n;
  }
  return r;
}
long long sqr(long long a) { return a * a; }
long long msqr(long long a) { return ((a % MOD) * (a % MOD)) % MOD; }
long long msub(long long a, long long b) { return ((a - b) % MOD + MOD) % MOD; }
long long madd(long long a, long long b) { return (a + b) % MOD; }
long long mmul(long long a, long long b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
int cal(int x, int s) {
  int cnt = 0;
  int len = (1 << s);
  return (x / len + 1) * (len / 2) -
         max(0, ((x / len) * len + (len / 2) - 1 - x));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int l, r;
    cin >> l >> r;
    int ans = INF;
    int len = floor(log2(r)) + 1;
    while (len) {
      int v1 = cal(r, len);
      int v2 = cal(l - 1, len);
      ans = min(ans, cal(r, len) - cal(l - 1, len));
      len--;
    }
    cout << ans << '\n';
  }
  return 0;
}

```