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>
using namespace std;
vector<long long> res;
bool isClassy(long long n) {
  int count = 0;
  while (n > 0) {
    if ((n % 10) != 0) count++;
    n = n / 10;
  }
  return count <= 3;
}
void precalc(vector<long long>& res, long long l, long long r) {
  for (long long i = l; i <= r; i++) {
    if (isClassy(i) == true) res.push_back(i);
  }
}
void brute(int pos, int count, long long curr) {
  if (pos == 18) {
    res.push_back(curr);
    return;
  }
  brute(pos + 1, count, curr * 10);
  if (count < 3) {
    for (int k = 1; k <= 9; k++) {
      brute(pos + 1, count + 1, curr * 10 + k);
    }
  }
}
int main() {
  brute(0, 0, 0);
  res.push_back(1000000000000000000);
  long long t;
  cin >> t;
  vector<pair<long long, long long>> q;
  long long l, r;
  while (t--) {
    cin >> l >> r;
    q.push_back({l, r});
  }
  t = q.size();
  for (int i = 0; i < t; i++) {
    long long L, R;
    L = q[i].first;
    R = q[i].second;
    cout << upper_bound(res.begin(), res.end(), R) -
                lower_bound(res.begin(), res.end(), L)
         << "\n";
  }
  return 0;
}