#include <bits/stdc++.h>
using namespace std;
int pw[4] = {1, 9, 81, 729};
long long com(long long n, long long k) {
  if (k == 0 || k == n)
    return 1;
  else if (k < 0 || k > n)
    return 0;
  else if (k == 1)
    return n;
  else {
    return com(n - 1, k - 1) + com(n - 1, k);
  }
}
long long get(long long n, long long k) {
  long long tot = 0;
  for (long long i = 0; i <= k; i++) {
    tot += com(n, i) * pw[i];
  }
  return tot;
}
long long cal(long long x) {
  stringstream ss;
  ss << x;
  string s = ss.str();
  long long len = s.length();
  int cnt = 3;
  long long res = 0;
  for (int i = 0; i < len; i++) {
    if (s[i] == '0')
      continue;
    else {
      res += get(len - i - 1, cnt);
      cnt--;
      if (cnt == -1) break;
      res += get(len - i - 1, cnt) * (s[i] - '1');
    }
  }
  return res;
}
int main() {
  int t;
  long long L, R;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> L >> R;
    cout << cal(R + 1) - cal(L) << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int pw[4] = {1, 9, 81, 729};
long long com(long long n, long long k) {
  if (k == 0 || k == n)
    return 1;
  else if (k < 0 || k > n)
    return 0;
  else if (k == 1)
    return n;
  else {
    return com(n - 1, k - 1) + com(n - 1, k);
  }
}
long long get(long long n, long long k) {
  long long tot = 0;
  for (long long i = 0; i <= k; i++) {
    tot += com(n, i) * pw[i];
  }
  return tot;
}
long long cal(long long x) {
  stringstream ss;
  ss << x;
  string s = ss.str();
  long long len = s.length();
  int cnt = 3;
  long long res = 0;
  for (int i = 0; i < len; i++) {
    if (s[i] == '0')
      continue;
    else {
      res += get(len - i - 1, cnt);
      cnt--;
      if (cnt == -1) break;
      res += get(len - i - 1, cnt) * (s[i] - '1');
    }
  }
  return res;
}
int main() {
  int t;
  long long L, R;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> L >> R;
    cout << cal(R + 1) - cal(L) << endl;
  }
  return 0;
}

```