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;
int Q;
long long l, r;
vector<long long> a;
void Find(long long cur, int s);
int main() {
  ios::sync_with_stdio(false);
  for (int i = 1; i <= 9; ++i) Find(i, 1);
  sort(a.begin(), a.end());
  cin >> Q;
  while (Q--) {
    cin >> l >> r;
    cout << upper_bound(a.begin(), a.end(), r) -
                lower_bound(a.begin(), a.end(), l)
         << endl;
  }
  return 0;
}
void Find(long long cur, int s) {
  a.push_back(cur);
  while (cur <= 1e17) {
    cur *= 10;
    a.push_back(cur);
    if (s < 3) {
      for (int i = 1; i < 10; i++) Find(cur + i, s + 1);
    }
  }
}