#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
void gen(long long at, long long cur, long long cnt) {
  if (at == 18) {
    v.push_back(cur);
    return;
  }
  gen(at + 1, cur * 10, cnt);
  if (cnt < 3)
    for (long long i = 1; i <= 9; i++) gen(at + 1, cur * 10 + i, cnt + 1);
}
int main() {
  gen(0, 0, 0);
  sort(v.begin(), v.end());
  v.push_back(1e18);
  long long i, j, t;
  long long l, r, n;
  cin >> t;
  while (t--) {
    cin >> l >> r;
    int R = upper_bound(v.begin(), v.end(), r) - v.begin();
    int L = lower_bound(v.begin(), v.end(), l) - v.begin();
    cout << R - L << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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> v;
void gen(long long at, long long cur, long long cnt) {
  if (at == 18) {
    v.push_back(cur);
    return;
  }
  gen(at + 1, cur * 10, cnt);
  if (cnt < 3)
    for (long long i = 1; i <= 9; i++) gen(at + 1, cur * 10 + i, cnt + 1);
}
int main() {
  gen(0, 0, 0);
  sort(v.begin(), v.end());
  v.push_back(1e18);
  long long i, j, t;
  long long l, r, n;
  cin >> t;
  while (t--) {
    cin >> l >> r;
    int R = upper_bound(v.begin(), v.end(), r) - v.begin();
    int L = lower_bound(v.begin(), v.end(), l) - v.begin();
    cout << R - L << endl;
  }
  return 0;
}

```