#include <bits/stdc++.h>
using namespace std;
int f(long long n, int k) {
  if (k == 0 || n == 0) return 0;
  if (n < 10) return n;
  return 9 + f(n / 10, k) + (9 - n % 10) * f(n / 10 - 1, k - 1) +
         n % 10 * f(n / 10, k - 1);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r, cout << f(r, 3) - f(l - 1, 3) << "\n";
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

Let's call some positive integer classy if its decimal representation contains no more than 3 non-zero digits. For example, numbers 4, 200000, 10203 are classy and numbers 4231, 102306, 7277420000 are not.

You are given a segment [L; R]. Count the number of classy integers x such that L ≤ x ≤ R.

Each testcase contains several segments, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer T (1 ≤ T ≤ 10^4) — the number of segments in a testcase.

Each of the next T lines contains two integers L_i and R_i (1 ≤ L_i ≤ R_i ≤ 10^{18}).

Output

Print T lines — the i-th line should contain the number of classy integers on a segment [L_i; R_i].

Example

Input

4
1 1000
1024 1024
65536 65536
999999 1000001


Output

1000
1
0
2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int f(long long n, int k) {
  if (k == 0 || n == 0) return 0;
  if (n < 10) return n;
  return 9 + f(n / 10, k) + (9 - n % 10) * f(n / 10 - 1, k - 1) +
         n % 10 * f(n / 10, k - 1);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r, cout << f(r, 3) - f(l - 1, 3) << "\n";
  }
}

```