#include <bits/stdc++.h>
using namespace std;
long long int sum() { return 0; }
template <typename T, typename... Args>
T sum(T a, Args... args) {
  return a + sum(args...);
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const long long int N = 200005;
vector<long long int> ans = {INF};
void precal(long long int pos, long long int c, long long int num) {
  if (pos == 18) {
    ans.push_back(num);
    return;
  }
  precal(pos + 1, c, num * 10);
  if (c < 3) {
    for (long long int i = 1; i <= 9; i++) {
      precal(pos + 1, c + 1, num * 10 + i);
    }
  }
}
void solve() {
  long long int l, r;
  cin >> l >> r;
  cout << upper_bound(ans.begin(), ans.end(), r) -
              lower_bound(ans.begin(), ans.end(), l)
       << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  precal(0, 0, 0);
  sort(ans.begin(), ans.end());
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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;
long long int sum() { return 0; }
template <typename T, typename... Args>
T sum(T a, Args... args) {
  return a + sum(args...);
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const long long int N = 200005;
vector<long long int> ans = {INF};
void precal(long long int pos, long long int c, long long int num) {
  if (pos == 18) {
    ans.push_back(num);
    return;
  }
  precal(pos + 1, c, num * 10);
  if (c < 3) {
    for (long long int i = 1; i <= 9; i++) {
      precal(pos + 1, c + 1, num * 10 + i);
    }
  }
}
void solve() {
  long long int l, r;
  cin >> l >> r;
  cout << upper_bound(ans.begin(), ans.end(), r) -
              lower_bound(ans.begin(), ans.end(), l)
       << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  precal(0, 0, 0);
  sort(ans.begin(), ans.end());
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```