#include <bits/stdc++.h>
using namespace std;
long long cnk[20][20];
int pow1[4];
long get(long long len, long long k) {
  long ans = 0;
  for (long long i = 0; i <= k; i++) {
    ans += cnk[len][i] * pow1[i];
  }
  return ans;
}
long long count(long long a) {
  string s = to_string(a);
  int cnt = 3;
  long long ans = 0;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '0') continue;
    ans += get(s.length() - i - 1, cnt);
    cnt--;
    if (cnt == -1) break;
    ans += get(s.length() - i - 1, cnt) * (s[i] - '1');
  }
  return ans;
}
int main() {
  pow1[0] = 1;
  pow1[1] = 9;
  pow1[2] = 81;
  pow1[3] = 729;
  for (int i = 0; i < 20; i++) {
    for (int j = 0; j < 20; j++) {
      if (j == 0 || i == j)
        cnk[i][j] = 1;
      else if (i > j) {
        cnk[i][j] = cnk[i - 1][j - 1] + cnk[i - 1][j];
      }
    }
  }
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    cout << count(r + 1) - count(l) << endl;
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 cnk[20][20];
int pow1[4];
long get(long long len, long long k) {
  long ans = 0;
  for (long long i = 0; i <= k; i++) {
    ans += cnk[len][i] * pow1[i];
  }
  return ans;
}
long long count(long long a) {
  string s = to_string(a);
  int cnt = 3;
  long long ans = 0;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '0') continue;
    ans += get(s.length() - i - 1, cnt);
    cnt--;
    if (cnt == -1) break;
    ans += get(s.length() - i - 1, cnt) * (s[i] - '1');
  }
  return ans;
}
int main() {
  pow1[0] = 1;
  pow1[1] = 9;
  pow1[2] = 81;
  pow1[3] = 729;
  for (int i = 0; i < 20; i++) {
    for (int j = 0; j < 20; j++) {
      if (j == 0 || i == j)
        cnk[i][j] = 1;
      else if (i > j) {
        cnk[i][j] = cnk[i - 1][j - 1] + cnk[i - 1][j];
      }
    }
  }
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    cout << count(r + 1) - count(l) << endl;
  }
  return 0;
}

```