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;
unsigned long long int ncr(unsigned long long int a, unsigned long long int b) {
  unsigned long long int ans = 1;
  for (unsigned long long int i = 1; i <= b; i++) {
    ans *= (a - i + 1);
  }
  for (unsigned long long int i = 1; i <= b; i++) {
    ans /= i;
  }
  return ans;
}
unsigned long long int power(unsigned long long int a,
                             unsigned long long int b) {
  unsigned long long int ans = 1;
  for (int i = 1; i <= b; i++) {
    ans *= a;
  }
  return ans;
}
unsigned long long int solvelessthansingle(int numofdigit) {
  unsigned long long int ans = ncr(numofdigit - 1, 1);
  ans *= 9;
  return ans;
}
unsigned long long int solvelessthandouble(int numofdigit) {
  unsigned long long int ans = ncr(numofdigit - 1, 2);
  ans *= 81;
  return ans;
}
unsigned long long int solvelessthantriple(unsigned long long int numofdigit) {
  unsigned long long int ans = ncr(numofdigit - 1, 3);
  ans *= 729;
  return ans;
}
unsigned long long int solvelessthannone(unsigned long long int numofdigit) {
  return 1;
}
unsigned long long int solve(unsigned long long int n,
                             unsigned long long int rem) {
  if (rem == 0) {
    return 0;
  }
  if (n < 10) {
    return n;
  }
  int numofdigit = 0;
  unsigned long long int n2 = n;
  while (n2) {
    numofdigit++;
    n2 /= 10;
  }
  unsigned long long int ans = 0;
  if (rem >= 1) ans += solvelessthansingle(numofdigit);
  if (rem >= 2) ans += solvelessthandouble(numofdigit);
  if (rem >= 3) ans += solvelessthantriple(numofdigit);
  int lastdigit;
  unsigned long long int n1 = n;
  while (n1) {
    int temp = n1 % 10;
    lastdigit = temp;
    n1 /= 10;
  }
  if (rem >= 1) ans += (lastdigit)*solvelessthannone(numofdigit);
  if (rem >= 2) ans += (lastdigit - 1) * solvelessthansingle(numofdigit);
  if (rem >= 3) ans += (lastdigit - 1) * solvelessthandouble(numofdigit);
  unsigned long long int mul =
      power((unsigned long long int)10, (unsigned long long int)numofdigit - 1);
  mul *= lastdigit;
  n -= mul;
  ans += solve(n, rem - 1);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    unsigned long long int n, m;
    cin >> n >> m;
    unsigned long long int ans = solve(m, 3) - solve(n - 1, 3);
    cout << ans << endl;
  }
  return 0;
}