#include <bits/stdc++.h>
using namespace std;
vector<long long int> v;
void call(long long int num, int len, int nonzero) {
  if (len == 18) {
    v.push_back(num);
    return;
  }
  call(num * 10, len + 1, nonzero);
  if (nonzero < 3) {
    for (int i = 1; i < 10; i++) {
      call(num * 10 + i, len + 1, nonzero + 1);
    }
  }
}
int main() {
  call(0, 0, 0);
  v.push_back(1000000000000000000);
  sort(v.begin(), v.end());
  int q;
  cin >> q;
  while (q--) {
    long long int a, b;
    scanf("%lld %lld", &a, &b);
    int cnt = (int)(upper_bound(v.begin(), v.end(), b) -
                    lower_bound(v.begin(), v.end(), a));
    printf("%d\n", cnt);
  }
  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;
vector<long long int> v;
void call(long long int num, int len, int nonzero) {
  if (len == 18) {
    v.push_back(num);
    return;
  }
  call(num * 10, len + 1, nonzero);
  if (nonzero < 3) {
    for (int i = 1; i < 10; i++) {
      call(num * 10 + i, len + 1, nonzero + 1);
    }
  }
}
int main() {
  call(0, 0, 0);
  v.push_back(1000000000000000000);
  sort(v.begin(), v.end());
  int q;
  cin >> q;
  while (q--) {
    long long int a, b;
    scanf("%lld %lld", &a, &b);
    int cnt = (int)(upper_bound(v.begin(), v.end(), b) -
                    lower_bound(v.begin(), v.end(), a));
    printf("%d\n", cnt);
  }
  return 0;
}

```