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;
long long com[20][20];
long long pow9(long long i) {
  long long mul = 1;
  while (i--) {
    mul *= 9;
  }
  return mul;
}
long long cal(long long x, long long choice) {
  string str = to_string(x);
  str = '0' + str;
  long long ans = 0, k = 0, m, d;
  for (int i = 0; i < str.size() - 1; i++) {
    if (str[i] != '0') k++;
    m = str.size() - (i + 2);
    if (str[i + 1] == '0') continue;
    d = str[i + 1] - '0';
    for (int j = 0; j <= 3 - k; j++) {
      ans += com[m][j] * pow9(j);
    }
    for (int j = 0; j <= 3 - k - 1; j++) {
      ans += com[m][j] * pow9(j) * (d - 1);
    }
  }
  if (str[str.size() - 1] != '0') k++;
  if (k <= 3 && choice == 1)
    return ans + 1;
  else
    return ans;
}
int main() {
  long long t, l, r;
  memset(com, 0, sizeof com);
  for (int i = 0; i < 20; i++) com[i][0] = 1, com[i][i] = 1;
  for (int i = 0; i < 20; i++) {
    for (int j = 0; j < i; j++) {
      if (com[i][j] == 0) com[i][j] = com[i - 1][j - 1] + com[i - 1][j];
    }
  }
  cin >> t;
  while (t--) {
    cin >> l >> r;
    cout << cal(r, 1) - cal(l, 2) << "\n";
  }
}