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

### 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;
vector<long long> v;
void dfs(long long u, int cnt, int step) {
  v.push_back(u);
  if (step == 18) return;
  dfs(u * 10, cnt, step + 1);
  if (cnt < 3) {
    for (int i = 1; i <= 9; i++) {
      dfs(u * 10 + i, cnt + 1, step + 1);
    }
  }
}
int main() {
  int q;
  for (int i = 1; i <= 9; i++) dfs(i, 1, 1);
  v.push_back(1e18);
  sort(v.begin(), v.end());
  scanf("%d", &q);
  while (q--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    l = lower_bound(v.begin(), v.end(), l) - v.begin();
    r = upper_bound(v.begin(), v.end(), r) - v.begin();
    cout << r - l << endl;
  }
}

```