#include <bits/stdc++.h>
using namespace std;
const int N = 500 + 10;
string s(18, '0');
vector<long long> ans;
void dfs(int cur, int cnt) {
  if (cur == 18) {
    ans.push_back(stoll(s));
    return;
  }
  if (cnt < 3) {
    for (char c = '1'; c <= '9'; c++) {
      s[cur] = c;
      dfs(cur + 1, cnt + 1);
    }
  }
  s[cur] = '0';
  dfs(cur + 1, cnt);
}
int main() {
  int t;
  cin >> t;
  dfs(0, 0);
  long long l, r;
  sort(ans.begin(), ans.end());
  ans.push_back((long long)1e18);
  while (t--) {
    cin >> l >> r;
    cout << upper_bound(ans.begin(), ans.end(), r) -
                lower_bound(ans.begin(), ans.end(), l)
         << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int N = 500 + 10;
string s(18, '0');
vector<long long> ans;
void dfs(int cur, int cnt) {
  if (cur == 18) {
    ans.push_back(stoll(s));
    return;
  }
  if (cnt < 3) {
    for (char c = '1'; c <= '9'; c++) {
      s[cur] = c;
      dfs(cur + 1, cnt + 1);
    }
  }
  s[cur] = '0';
  dfs(cur + 1, cnt);
}
int main() {
  int t;
  cin >> t;
  dfs(0, 0);
  long long l, r;
  sort(ans.begin(), ans.end());
  ans.push_back((long long)1e18);
  while (t--) {
    cin >> l >> r;
    cout << upper_bound(ans.begin(), ans.end(), r) -
                lower_bound(ans.begin(), ans.end(), l)
         << endl;
  }
  return 0;
}

```