#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 20;
const long long INFLL = 2e18 + 20;
const long double pi = 3.14159265358979323846;
const long double EPSILION = 0.000000000001;
const int MOD = 1e9 + 7;
const int MAX = 2e5 + 20;
void usaco(string filename) {
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
void init() {}
long long getCount(long long n, long long k) {
  long long res = (n >> (k + 1)) << k;
  if ((n >> k) & 1) {
    res += n & ((1LL << k) - 1);
  }
  return res;
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long ans = r - l + 1;
  for (auto i = 0; i < 30; i++) {
    long long x = getCount(l, i);
    long long y = getCount(r + 1, i);
    ans = min(ans, (r - l + 1) - (y - x));
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  init();
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
}

### 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 INF = 1e9 + 20;
const long long INFLL = 2e18 + 20;
const long double pi = 3.14159265358979323846;
const long double EPSILION = 0.000000000001;
const int MOD = 1e9 + 7;
const int MAX = 2e5 + 20;
void usaco(string filename) {
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
void init() {}
long long getCount(long long n, long long k) {
  long long res = (n >> (k + 1)) << k;
  if ((n >> k) & 1) {
    res += n & ((1LL << k) - 1);
  }
  return res;
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long ans = r - l + 1;
  for (auto i = 0; i < 30; i++) {
    long long x = getCount(l, i);
    long long y = getCount(r + 1, i);
    ans = min(ans, (r - l + 1) - (y - x));
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  init();
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
}

```