#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long extgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
long long revrsno(long long n) {
  long long k = 0;
  vector<long long> z;
  while (n > 0) {
    z.push_back(n % 10);
    n /= 10;
  }
  for (long long i = 0; i < z.size(); i++) k = k * 10 + z[i];
  return k;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long divide(long long a, long long b) {
  return (a % 1000000007) * powm(b, 1000000007 - 2) % 1000000007;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long add(long long a, long long b) {
  return (a % 1000000007 + b % 1000000007) % 1000000007;
}
long long Totfun(long long n) {
  long long z = n;
  if (n % 2 == 0) {
    while (n % 2 == 0) {
      n /= 2;
    }
    z /= 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (isPrime(i) && n % i == 0) {
      while (n % i == 0) {
        n /= i;
      }
      z -= z / i;
    }
  }
  if (n > 1) {
    z -= z / n;
  }
  return z;
}
long long maxPrimeFactor(long long n) {
  long long maxPrime = -1;
  while (n % 2 == 0) {
    maxPrime = 2;
    n >>= 1;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    while (n % i == 0) {
      maxPrime = i;
      n = n / i;
    }
  }
  if (n > 2) maxPrime = n;
  return maxPrime;
}
long long fact[100001];
long long nCr(long long n, long long r) {
  if (r > n || r < 0) return 0;
  long long z = fact[n];
  z = mul(z, powm(fact[n - r], 1000000007 - 2));
  z = mul(z, powm(fact[r], 1000000007 - 2));
  return z;
}
long long v[200001][19];
void cal() {
  for (long long i = 0; i < 19; i++) {
    for (long long j = 1; j < 200001; j++) {
      if ((j & (1 << i)))
        v[j][i] = 1;
      else
        v[j][i] = 0;
    }
  }
  for (long long i = 0; i < 19; i++) {
    for (long long j = 1; j < 200001; j++) {
      v[j][i] += v[j - 1][i];
    }
  }
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long ans = 1e9;
  for (long long i = 0; i < 19; i++) {
    ans = ((ans > (r - l + 1) - (v[r][i] - (v[l - 1][i])))
               ? (r - l + 1) - (v[r][i] - (v[l - 1][i]))
               : ans);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  cal();
  while (t--) {
    solve();
    cout << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long extgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
long long revrsno(long long n) {
  long long k = 0;
  vector<long long> z;
  while (n > 0) {
    z.push_back(n % 10);
    n /= 10;
  }
  for (long long i = 0; i < z.size(); i++) k = k * 10 + z[i];
  return k;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long divide(long long a, long long b) {
  return (a % 1000000007) * powm(b, 1000000007 - 2) % 1000000007;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long add(long long a, long long b) {
  return (a % 1000000007 + b % 1000000007) % 1000000007;
}
long long Totfun(long long n) {
  long long z = n;
  if (n % 2 == 0) {
    while (n % 2 == 0) {
      n /= 2;
    }
    z /= 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (isPrime(i) && n % i == 0) {
      while (n % i == 0) {
        n /= i;
      }
      z -= z / i;
    }
  }
  if (n > 1) {
    z -= z / n;
  }
  return z;
}
long long maxPrimeFactor(long long n) {
  long long maxPrime = -1;
  while (n % 2 == 0) {
    maxPrime = 2;
    n >>= 1;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    while (n % i == 0) {
      maxPrime = i;
      n = n / i;
    }
  }
  if (n > 2) maxPrime = n;
  return maxPrime;
}
long long fact[100001];
long long nCr(long long n, long long r) {
  if (r > n || r < 0) return 0;
  long long z = fact[n];
  z = mul(z, powm(fact[n - r], 1000000007 - 2));
  z = mul(z, powm(fact[r], 1000000007 - 2));
  return z;
}
long long v[200001][19];
void cal() {
  for (long long i = 0; i < 19; i++) {
    for (long long j = 1; j < 200001; j++) {
      if ((j & (1 << i)))
        v[j][i] = 1;
      else
        v[j][i] = 0;
    }
  }
  for (long long i = 0; i < 19; i++) {
    for (long long j = 1; j < 200001; j++) {
      v[j][i] += v[j - 1][i];
    }
  }
}
void solve() {
  long long l, r;
  cin >> l >> r;
  long long ans = 1e9;
  for (long long i = 0; i < 19; i++) {
    ans = ((ans > (r - l + 1) - (v[r][i] - (v[l - 1][i])))
               ? (r - l + 1) - (v[r][i] - (v[l - 1][i]))
               : ans);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  cal();
  while (t--) {
    solve();
    cout << endl;
  }
  return 0;
}

```