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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int64_t f(int64_t x) {
  if (x < 1000) return x;
  std::vector<int> s;
  while (x) {
    s.push_back(x % 10);
    x /= 10;
  }
  reverse(s.begin(), s.end());
  int len = static_cast<int>(s.size()) - 1;
  int64_t res = 0;
  for (int i = 1; i != len + 1; ++i) {
    res += 9 + 9 * 9 * (i - 1) + 9 * 9 * 9 * (i - 1) * (i - 2) / 2;
  }
  ++len;
  res +=
      (s[0] - 1) * (1 + 9 * (len - 1) + 9 * 9 * (len - 1) * (len - 2) / 2) + 1;
  int i = 1;
  for (; i != static_cast<int>(s.size()); ++i) {
    if (s[i] != 0) break;
  }
  len = static_cast<int>(s.size()) - i;
  if (len == 0) return res;
  res += s[i];
  if (len != 1) {
    res += (s[i] - 1) * 9 * (len - 1) + 9 * 9 * (len - 1) * (len - 2) / 2 +
           9 * (len - 1);
    bool t = false;
    ++i;
    for (; i != static_cast<int>(s.size()); ++i) {
      res += (t ? 9 : s[i]);
      if (s[i] != 0) t = true;
    }
  }
  return res;
}
int solve(int64_t l, int64_t r) { return f(r) - f(l - 1); }
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  int t;
  std::cin >> t;
  while (t--) {
    int64_t l, r;
    std::cin >> l >> r;
    std::cout << solve(l, r) << '\n';
  }
  return 0;
}