#include <bits/stdc++.h>
using namespace std;
long long a[100], dis[100][50], l, r, num, ans, down, tot;
long long dfs(long long book, long long go, bool to) {
  if (book == -1) {
    return 1;
  }
  if (to == 0 && dis[book][go] != -1) {
    return dis[book][go];
  }
  long long down = to ? a[book] : 9;
  long long ans = 0;
  for (long long i = 0; i <= down; i++) {
    if (i == 0) {
      ans = ans + dfs(book - 1, go, to && a[book] == i);
    } else if (go != 3) {
      ans = ans + dfs(book - 1, go + 1, to && a[book] == i);
    }
  }
  if (to == 0) {
    dis[book][go] = ans;
  }
  return ans;
}
long long run(long long x) {
  tot = 0;
  while (x > 0) {
    a[tot++] = x % 10;
    x = x / 10;
  }
  return dfs(tot - 1, 0, true);
}
int main() {
  memset(dis, -1, sizeof(dis));
  cin >> num;
  while (num--) {
    cin >> l >> r;
    cout << run(r) - run(l - 1) << endl;
  }
  return 0;
}

### Prompt

Please formulate 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;
long long a[100], dis[100][50], l, r, num, ans, down, tot;
long long dfs(long long book, long long go, bool to) {
  if (book == -1) {
    return 1;
  }
  if (to == 0 && dis[book][go] != -1) {
    return dis[book][go];
  }
  long long down = to ? a[book] : 9;
  long long ans = 0;
  for (long long i = 0; i <= down; i++) {
    if (i == 0) {
      ans = ans + dfs(book - 1, go, to && a[book] == i);
    } else if (go != 3) {
      ans = ans + dfs(book - 1, go + 1, to && a[book] == i);
    }
  }
  if (to == 0) {
    dis[book][go] = ans;
  }
  return ans;
}
long long run(long long x) {
  tot = 0;
  while (x > 0) {
    a[tot++] = x % 10;
    x = x / 10;
  }
  return dfs(tot - 1, 0, true);
}
int main() {
  memset(dis, -1, sizeof(dis));
  cin >> num;
  while (num--) {
    cin >> l >> r;
    cout << run(r) - run(l - 1) << endl;
  }
  return 0;
}

```