#include <bits/stdc++.h>
using namespace std;
vector<long long> x;
void rec(long long val, int len, int cnt) {
  if (cnt == 4) return;
  if (len == 18) {
    x.push_back(val);
    return;
  }
  for (int i = 0; i < 10; i++) rec(val * 10 + i, len + 1, cnt + (i != 0));
}
int main() {
  int t;
  cin >> t;
  rec(0, 0, 0);
  x.push_back((long long)1e18);
  while (t--) {
    long long l, r;
    cin >> l >> r;
    long long ans = upper_bound(x.begin(), x.end(), r) -
                    upper_bound(x.begin(), x.end(), l - 1);
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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> x;
void rec(long long val, int len, int cnt) {
  if (cnt == 4) return;
  if (len == 18) {
    x.push_back(val);
    return;
  }
  for (int i = 0; i < 10; i++) rec(val * 10 + i, len + 1, cnt + (i != 0));
}
int main() {
  int t;
  cin >> t;
  rec(0, 0, 0);
  x.push_back((long long)1e18);
  while (t--) {
    long long l, r;
    cin >> l >> r;
    long long ans = upper_bound(x.begin(), x.end(), r) -
                    upper_bound(x.begin(), x.end(), l - 1);
    cout << ans << endl;
  }
  return 0;
}

```