#include <bits/stdc++.h>
using namespace std;
string num;
long long memo[2][20][4];
long long go(bool lt = false, int n = 0, int nz = 0) {
  if (nz >= 4) return 0;
  if (n >= int((num).size())) return 1;
  long long& res = memo[lt][n][nz];
  if (res < 0) {
    res = 0;
    for (int d = (0), _b = (lt ? 9 : num[n] - '0'); d <= _b; ++d) {
      res += go(lt || d < num[n] - '0', n + 1, nz + (d > 0));
    }
  }
  return res;
}
long long F(long long R) {
  num = to_string(R);
  memset(memo, -1, sizeof(memo));
  return go();
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int Q;
  cin >> Q;
  for (int q = (1), _b = (Q); q <= _b; ++q) {
    long long L, R;
    cin >> L >> R;
    long long res = F(R) - F(L - 1);
    cout << res << '\n';
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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;
string num;
long long memo[2][20][4];
long long go(bool lt = false, int n = 0, int nz = 0) {
  if (nz >= 4) return 0;
  if (n >= int((num).size())) return 1;
  long long& res = memo[lt][n][nz];
  if (res < 0) {
    res = 0;
    for (int d = (0), _b = (lt ? 9 : num[n] - '0'); d <= _b; ++d) {
      res += go(lt || d < num[n] - '0', n + 1, nz + (d > 0));
    }
  }
  return res;
}
long long F(long long R) {
  num = to_string(R);
  memset(memo, -1, sizeof(memo));
  return go();
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int Q;
  cin >> Q;
  for (int q = (1), _b = (Q); q <= _b; ++q) {
    long long L, R;
    cin >> L >> R;
    long long res = F(R) - F(L - 1);
    cout << res << '\n';
  }
  return 0;
}

```