#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
constexpr ll MAX = 2000000000000000000;
constexpr ld PI = 3.14159265358979;
constexpr ll MOD = 0;
ld dotorad(ld K) { return PI * K / 180.0; }
ld radtodo(ld K) { return K * 180.0 / PI; }
mt19937 mt;
void randinit() {
  srand((unsigned)time(NULL));
  mt = mt19937(rand());
}
ll c(ll N, ll i) {
  N++;
  ll p = (1LL << i), q = (1LL << (i + 1));
  ll ans = N / q * p;
  N %= q;
  ans += min(N, p);
  return ans;
}
int main() {
  ll T;
  cin >> T;
  for (ll t = 0; t < T; t++) {
    ll L, R;
    cin >> L >> R;
    ll ans = MAX;
    for (ll i = 0; i < 20; i++) {
      ans = min(ans, c(R, i) - c(L - 1, i));
    }
    cout << ans << endl;
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
constexpr ll MAX = 2000000000000000000;
constexpr ld PI = 3.14159265358979;
constexpr ll MOD = 0;
ld dotorad(ld K) { return PI * K / 180.0; }
ld radtodo(ld K) { return K * 180.0 / PI; }
mt19937 mt;
void randinit() {
  srand((unsigned)time(NULL));
  mt = mt19937(rand());
}
ll c(ll N, ll i) {
  N++;
  ll p = (1LL << i), q = (1LL << (i + 1));
  ll ans = N / q * p;
  N %= q;
  ans += min(N, p);
  return ans;
}
int main() {
  ll T;
  cin >> T;
  for (ll t = 0; t < T; t++) {
    ll L, R;
    cin >> L >> R;
    ll ans = MAX;
    for (ll i = 0; i < 20; i++) {
      ans = min(ans, c(R, i) - c(L - 1, i));
    }
    cout << ans << endl;
  }
}

```