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 t, l = 1;
unsigned long long lft, rgt;
unsigned long long numb[1000000];
queue<pair<unsigned long long, int> > que;
int binS(unsigned long long val);
void gen(unsigned long long nr, int cnt = 0);
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  for (int i = 1; i < 10; ++i) que.push({i, 1});
  while (!que.empty()) {
    numb[++l] = que.front().first;
    int cnt = que.front().second;
    que.pop();
    if (numb[l] * 10ULL <= (unsigned long long)1000000000000000000ULL)
      que.push({numb[l] * 10ULL, cnt});
    if (numb[l] * 10ULL + 1ULL <= (unsigned long long)1000000000000000000ULL &&
        cnt + 1 <= 3) {
      for (int i = 1; i <= 9; ++i)
        que.push({numb[l] * 10ULL + i * 1ULL, cnt + 1});
    }
  }
  cin >> t;
  while (t--) {
    cin >> lft >> rgt;
    cout << binS(rgt) - binS(lft - 1) << "\n";
  }
  return 0;
}
int binS(unsigned long long val) {
  if (val == (unsigned long long)1000000000000000000ULL) return l;
  int s = 1, f = l, m;
  while (s < f) {
    m = (s + f) / 2;
    if (numb[m] <= val)
      s = m + 1;
    else
      f = m;
  }
  return s - 1;
}