#include <bits/stdc++.h>
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 5;
const int mo = 1e9 + 7;
using namespace std;
std::vector<std::vector<int>> sum(N, std::vector<int>(22, 0));
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
void Pre() {
  for (int i = 1; i <= N - 5; i++)
    for (int j = 0; j <= 20; j++) sum[i][j] = sum[i - 1][j] + ((i >> j) & 1);
}
int main() {
  int T, l, r;
  read(T);
  Pre();
  while (T--) {
    read(l), read(r);
    int ans = 0;
    l--;
    for (int i = 0; i <= 20; i++) ans = max(ans, sum[r][i] - sum[l][i]);
    writeln(r - l - ans);
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 5;
const int mo = 1e9 + 7;
using namespace std;
std::vector<std::vector<int>> sum(N, std::vector<int>(22, 0));
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
void Pre() {
  for (int i = 1; i <= N - 5; i++)
    for (int j = 0; j <= 20; j++) sum[i][j] = sum[i - 1][j] + ((i >> j) & 1);
}
int main() {
  int T, l, r;
  read(T);
  Pre();
  while (T--) {
    read(l), read(r);
    int ans = 0;
    l--;
    for (int i = 0; i <= 20; i++) ans = max(ans, sum[r][i] - sum[l][i]);
    writeln(r - l - ans);
  }
}

```