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 DP[20][4];
void init() {
  DP[0][3] = DP[0][2] = DP[0][1] = DP[0][0] = 1;
  for (int i = 1; i <= 18; i++) {
    DP[i][0] = 1;
    for (int j = 1; j <= 3; j++) {
      DP[i][j] = 9 * DP[i - 1][j - 1] + DP[i - 1][j];
    }
  }
}
long long sol(int i, int j, string str) {
  if (i == 0) {
    if (j >= 0)
      return 1;
    else
      return 0;
  }
  int d = str[i - 1] - '0';
  if (d == 0)
    return sol(i - 1, j, str);
  else if (j == 0)
    return 1;
  return DP[i - 1][j] + (d - 1) * DP[i - 1][j - 1] + sol(i - 1, j - 1, str);
}
string tostr(long long x) {
  string rtn;
  while (x) {
    rtn += (x % 10 + '0');
    x /= 10;
  }
  return rtn;
}
string A, B;
long long a, b;
int cas;
int main() {
  ios::sync_with_stdio(false);
  init();
  cin >> cas;
  while (cas--) {
    cin >> a >> b;
    a--;
    A = tostr(a);
    B = tostr(b);
    cout << sol(B.length(), 3, B) - sol(A.length(), 3, A) << '\n';
  }
  return 0;
}