#include <bits/stdc++.h>
using namespace std;
vector<long long int> all;
void go(int on, long long int cur, int qnt) {
  if (on == 0) {
    if (cur > 0) all.push_back(cur);
    return;
  }
  go(on - 1, 10ll * cur, qnt);
  if (qnt == 0) return;
  for (int i = 1; i <= 9; i++) {
    go(on - 1, 10ll * cur + i, qnt - 1);
  }
}
int main() {
  long long int inf = 1e18;
  go(18, 0, 3);
  all.push_back(inf);
  sort(all.begin(), all.end());
  all.resize(unique(all.begin(), all.end()) - all.begin());
  int t;
  cin >> t;
  while (t--) {
    long long int l, r;
    cin >> l >> r;
    int q = upper_bound(all.begin(), all.end(), r) -
            lower_bound(all.begin(), all.end(), l);
    cout << q << '\n';
  }
  return 0;
}

### Prompt

Generate 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;
vector<long long int> all;
void go(int on, long long int cur, int qnt) {
  if (on == 0) {
    if (cur > 0) all.push_back(cur);
    return;
  }
  go(on - 1, 10ll * cur, qnt);
  if (qnt == 0) return;
  for (int i = 1; i <= 9; i++) {
    go(on - 1, 10ll * cur + i, qnt - 1);
  }
}
int main() {
  long long int inf = 1e18;
  go(18, 0, 3);
  all.push_back(inf);
  sort(all.begin(), all.end());
  all.resize(unique(all.begin(), all.end()) - all.begin());
  int t;
  cin >> t;
  while (t--) {
    long long int l, r;
    cin >> l >> r;
    int q = upper_bound(all.begin(), all.end(), r) -
            lower_bound(all.begin(), all.end(), l);
    cout << q << '\n';
  }
  return 0;
}

```