#include <bits/stdc++.h>
using namespace std;
long long n, m;
vector<long long> ve;
void dfs(long long num, long long len, long long sum) {
  ve.push_back(num);
  if (len == 19) return;
  dfs(num * 10 + 0, len + 1, sum);
  if (sum < 3) {
    for (long long i = 1; i <= 9; i++) {
      dfs(num * 10 + i, len + 1, sum + 1);
    }
  }
}
signed main() {
  for (long long i = 1; i <= 9; i++) {
    dfs(i, 1, 1);
  }
  sort(ve.begin(), ve.end());
  long long T;
  cin >> T;
  while (T--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    cout << upper_bound(ve.begin(), ve.end(), r) -
                lower_bound(ve.begin(), ve.end(), l)
         << 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 n, m;
vector<long long> ve;
void dfs(long long num, long long len, long long sum) {
  ve.push_back(num);
  if (len == 19) return;
  dfs(num * 10 + 0, len + 1, sum);
  if (sum < 3) {
    for (long long i = 1; i <= 9; i++) {
      dfs(num * 10 + i, len + 1, sum + 1);
    }
  }
}
signed main() {
  for (long long i = 1; i <= 9; i++) {
    dfs(i, 1, 1);
  }
  sort(ve.begin(), ve.end());
  long long T;
  cin >> T;
  while (T--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    cout << upper_bound(ve.begin(), ve.end(), r) -
                lower_bound(ve.begin(), ve.end(), l)
         << endl;
  }
  return 0;
}

```