#include <bits/stdc++.h>
using namespace std;
const unsigned long long lim = 1000000000000000000;
vector<unsigned long long> v;
void enumerate(unsigned long long d, int c) {
  v.push_back(d);
  if (d * 10 <= lim) enumerate(d * 10, c);
  for (int i = 1; i <= 9; ++i) {
    if (d * 10 + i <= lim && c + 1 <= 3) {
      enumerate(d * 10 + i, c + 1);
    }
  }
}
int main() {
  for (int i = 1; i <= 9; ++i) {
    enumerate(i, 1);
  }
  sort(v.begin(), v.end());
  int Q;
  cin >> Q;
  while (Q--) {
    unsigned long long L, R;
    cin >> L >> R;
    cout << (lower_bound(v.begin(), v.end(), R + 1) -
             lower_bound(v.begin(), v.end(), L))
         << endl;
  }
  return 0;
}

### 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;
const unsigned long long lim = 1000000000000000000;
vector<unsigned long long> v;
void enumerate(unsigned long long d, int c) {
  v.push_back(d);
  if (d * 10 <= lim) enumerate(d * 10, c);
  for (int i = 1; i <= 9; ++i) {
    if (d * 10 + i <= lim && c + 1 <= 3) {
      enumerate(d * 10 + i, c + 1);
    }
  }
}
int main() {
  for (int i = 1; i <= 9; ++i) {
    enumerate(i, 1);
  }
  sort(v.begin(), v.end());
  int Q;
  cin >> Q;
  while (Q--) {
    unsigned long long L, R;
    cin >> L >> R;
    cout << (lower_bound(v.begin(), v.end(), R + 1) -
             lower_bound(v.begin(), v.end(), L))
         << endl;
  }
  return 0;
}

```