#include <bits/stdc++.h>
using namespace std;
const long long iinf = 1e9 + 10;
const long long inf = 1ll << 60;
const long long mod = 1e9 + 7;
void GG() {
  cout << "0\n";
  exit(0);
}
long long mpow(long long a, long long n, long long mo = mod) {
  long long re = 1;
  while (n > 0) {
    if (n & 1) re = re * a % mo;
    a = a * a % mo;
    n >>= 1;
  }
  return re;
}
long long inv(long long b, long long mo = mod) {
  if (b == 1) return b;
  return (mo - mo / b) * inv(mo % b, mo) % mo;
}
const long long maxn = 2e5 + 5;
long long cnt[20][maxn];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  for (long long i = 0; i < 20; ++i) {
    for (long long j = 1; j <= maxn - 1; ++j) {
      cnt[i][j] = cnt[i][j - 1] + ((j & (1 << i)) != 0);
    }
  }
  long long t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    long long re = 0;
    for (long long i = 0; i < 20; ++i) {
      re = max(re, (__typeof__(re))(cnt[i][r] - cnt[i][l - 1]));
    }
    cout << (r - l + 1) - re << '\n';
  }
}

### 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 long long iinf = 1e9 + 10;
const long long inf = 1ll << 60;
const long long mod = 1e9 + 7;
void GG() {
  cout << "0\n";
  exit(0);
}
long long mpow(long long a, long long n, long long mo = mod) {
  long long re = 1;
  while (n > 0) {
    if (n & 1) re = re * a % mo;
    a = a * a % mo;
    n >>= 1;
  }
  return re;
}
long long inv(long long b, long long mo = mod) {
  if (b == 1) return b;
  return (mo - mo / b) * inv(mo % b, mo) % mo;
}
const long long maxn = 2e5 + 5;
long long cnt[20][maxn];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  for (long long i = 0; i < 20; ++i) {
    for (long long j = 1; j <= maxn - 1; ++j) {
      cnt[i][j] = cnt[i][j - 1] + ((j & (1 << i)) != 0);
    }
  }
  long long t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    long long re = 0;
    for (long long i = 0; i < 20; ++i) {
      re = max(re, (__typeof__(re))(cnt[i][r] - cnt[i][l - 1]));
    }
    cout << (r - l + 1) - re << '\n';
  }
}

```