#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = double;
using pii = pair<int, int>;
using vec = vector<int>;
using veg = vector<pii>;
template <typename T>
T read() {
  T x = 0;
  bool f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return f ? -x : x;
}
template <typename T>
void chkmax(T &x, const T &y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, const T &y) {
  x = x < y ? x : y;
}
const int N = 3e5 + 10;
const int V = 2e5 + 10;
int s[25][N];
void solve() {
  int l = read<int>(), r = read<int>(), ans = 0;
  for (int i = (0); i <= (20); i++) {
    chkmax(ans, s[i][r] - s[i][l - 1]);
  }
  printf("%d\n", r - l + 1 - ans);
}
int main() {
  for (int i = (0); i <= (20); i++)
    for (int j = (1); j <= (V); j++) s[i][j] = s[i][j - 1] + (j >> i & 1);
  int T = read<int>();
  while (T--) solve();
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
using db = double;
using pii = pair<int, int>;
using vec = vector<int>;
using veg = vector<pii>;
template <typename T>
T read() {
  T x = 0;
  bool f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return f ? -x : x;
}
template <typename T>
void chkmax(T &x, const T &y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, const T &y) {
  x = x < y ? x : y;
}
const int N = 3e5 + 10;
const int V = 2e5 + 10;
int s[25][N];
void solve() {
  int l = read<int>(), r = read<int>(), ans = 0;
  for (int i = (0); i <= (20); i++) {
    chkmax(ans, s[i][r] - s[i][l - 1]);
  }
  printf("%d\n", r - l + 1 - ans);
}
int main() {
  for (int i = (0); i <= (20); i++)
    for (int j = (1); j <= (V); j++) s[i][j] = s[i][j - 1] + (j >> i & 1);
  int T = read<int>();
  while (T--) solve();
  return 0;
}

```