#include <bits/stdc++.h>
using namespace std;
long long dirs8[] = {0, -1, -1, 0, 1, -1, 1, 1, 0};
long long dirs4[] = {0, 1, 0, -1, 0};
long long dirs2[] = {0, 1, 0};
long long fact(long long n) {
  long long ans = 1;
  while (n) ans *= n--;
  return ans;
}
bool isPrime(long long &n) {
  for (long long i = 2; i * i <= n; ++i)
    if (n % i == 0) return 0;
  return 1;
}
vector<pair<long long, long long>> getPrimeFactors(long long n) {
  vector<pair<long long, long long>> facts;
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      long long cnt = 0;
      while (n % i == 0) {
        ++cnt;
        n /= i;
      }
      facts.push_back({i, cnt});
    }
  }
  if (n > 1) facts.push_back({n, 1});
  return facts;
}
const long long N = 200005;
long long dp[N][32];
void solve() {
  long long low, high;
  cin >> low >> high;
  low--;
  long long res = INT_MAX;
  for (long long i = 0; i < 32; i++) {
    res = min(res, high - low - dp[high][i] + dp[low][i]);
  }
  cout << res << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(dp[0], 0, sizeof dp[0]);
  for (long long i = 1; i <= N; i++) {
    for (long long c = 0; c < 32; c++) {
      dp[i][c] = dp[i - 1][c];
    }
    bitset<19> b(i);
    for (long long j = 0; j < 32; j++) {
      if (b[j] == true) {
        dp[i][j]++;
      }
    }
  }
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
long long dirs8[] = {0, -1, -1, 0, 1, -1, 1, 1, 0};
long long dirs4[] = {0, 1, 0, -1, 0};
long long dirs2[] = {0, 1, 0};
long long fact(long long n) {
  long long ans = 1;
  while (n) ans *= n--;
  return ans;
}
bool isPrime(long long &n) {
  for (long long i = 2; i * i <= n; ++i)
    if (n % i == 0) return 0;
  return 1;
}
vector<pair<long long, long long>> getPrimeFactors(long long n) {
  vector<pair<long long, long long>> facts;
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      long long cnt = 0;
      while (n % i == 0) {
        ++cnt;
        n /= i;
      }
      facts.push_back({i, cnt});
    }
  }
  if (n > 1) facts.push_back({n, 1});
  return facts;
}
const long long N = 200005;
long long dp[N][32];
void solve() {
  long long low, high;
  cin >> low >> high;
  low--;
  long long res = INT_MAX;
  for (long long i = 0; i < 32; i++) {
    res = min(res, high - low - dp[high][i] + dp[low][i]);
  }
  cout << res << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(dp[0], 0, sizeof dp[0]);
  for (long long i = 1; i <= N; i++) {
    for (long long c = 0; c < 32; c++) {
      dp[i][c] = dp[i - 1][c];
    }
    bitset<19> b(i);
    for (long long j = 0; j < 32; j++) {
      if (b[j] == true) {
        dp[i][j]++;
      }
    }
  }
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```