#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
void dfs(int depth, long long num, int count_) {
  v.push_back(num);
  if (depth == 18) return;
  dfs(depth + 1, num * 10, count_);
  if (count_ < 3) {
    for (int i = 1; i <= 9; i++) dfs(depth + 1, num * 10 + i, count_ + 1);
  }
}
int main() {
  for (int i = 1; i <= 9; i++) dfs(1, i, 1);
  v.push_back(1e18);
  sort(v.begin(), v.end());
  int n;
  scanf("%d", &n);
  while (n--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    long long sum =
        upper_bound(v.begin(), v.end(), r) - lower_bound(v.begin(), v.end(), l);
    printf("%lld\n", sum);
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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> v;
void dfs(int depth, long long num, int count_) {
  v.push_back(num);
  if (depth == 18) return;
  dfs(depth + 1, num * 10, count_);
  if (count_ < 3) {
    for (int i = 1; i <= 9; i++) dfs(depth + 1, num * 10 + i, count_ + 1);
  }
}
int main() {
  for (int i = 1; i <= 9; i++) dfs(1, i, 1);
  v.push_back(1e18);
  sort(v.begin(), v.end());
  int n;
  scanf("%d", &n);
  while (n--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    long long sum =
        upper_bound(v.begin(), v.end(), r) - lower_bound(v.begin(), v.end(), l);
    printf("%lld\n", sum);
  }
}

```