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 int> v;
void build(long long int n, long long int no, int p) {
  if (p > 18) {
    v.push_back(n);
    return;
  }
  build(n * 10, no, p + 1);
  if (no < 3) {
    for (int i = 1; i <= 9; i++) {
      build(n * 10 + i, no + 1, p + 1);
    }
  }
}
void solve() {
  long long int l, r;
  cin >> l >> r;
  cout << upper_bound(v.begin(), v.end(), r) -
              lower_bound(v.begin(), v.end(), l)
       << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  build(0, 0, 1);
  v.push_back(1000000000000000000);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}