#include <bits/stdc++.h>
using namespace std;
void in(int &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while ((c > '9' || c < '0') && c != '-') c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  x *= f;
}
int t, n, m, a[200005];
int s[200005][25], sum[200005][25];
signed main() {
  in(t);
  for (int i = 0; i <= 200000; i++) {
    int k = 0, m = i;
    while (m) {
      if (m % 2) s[i][k] = 1;
      m /= 2;
      k++;
    }
  }
  for (int i = 1; i <= 200000; i++) {
    for (int j = 0; j <= 20; j++) {
      s[i][j] += s[i - 1][j];
    }
  }
  while (t--) {
    in(n);
    in(m);
    int ans = 0;
    for (int i = 0; i <= 20; i++) {
      if (s[m][i] - s[n - 1][i]) ans = max(ans, s[m][i] - s[n - 1][i]);
    }
    printf("%d\n", m - n + 1 - ans);
  }
}

### 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;
void in(int &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while ((c > '9' || c < '0') && c != '-') c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  x *= f;
}
int t, n, m, a[200005];
int s[200005][25], sum[200005][25];
signed main() {
  in(t);
  for (int i = 0; i <= 200000; i++) {
    int k = 0, m = i;
    while (m) {
      if (m % 2) s[i][k] = 1;
      m /= 2;
      k++;
    }
  }
  for (int i = 1; i <= 200000; i++) {
    for (int j = 0; j <= 20; j++) {
      s[i][j] += s[i - 1][j];
    }
  }
  while (t--) {
    in(n);
    in(m);
    int ans = 0;
    for (int i = 0; i <= 20; i++) {
      if (s[m][i] - s[n - 1][i]) ans = max(ans, s[m][i] - s[n - 1][i]);
    }
    printf("%d\n", m - n + 1 - ans);
  }
}

```