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 count_classy(long long N) {
  if (N == 0) return 0;
  string S = to_string(N);
  long long res = 0;
  for (int l = 1; l < S.size(); l++) {
    res += 9 * (1 + 9 * (l - 1) + 9 * 9 * (l - 1) * (l - 2) / 2);
  }
  int j = 3;
  for (int i = 0; i < S.size(); i++) {
    if (S[i] == '0') continue;
    int l = S.size() - i;
    if (i > 0) {
      if (j >= 0) res += 1;
      if (j >= 1) res += 9 * (l - 1);
      if (j >= 2) res += 9 * 9 * (l - 1) * (l - 2) / 2;
    }
    int d = S[i] - '0' - 1;
    j--;
    if (j < 0) break;
    if (j >= 0) res += d;
    if (j >= 1) res += d * 9 * (l - 1);
    if (j >= 2) res += d * 9 * 9 * (l - 1) * (l - 2) / 2;
  }
  if (j >= 0) res++;
  return res;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    long long L, R;
    scanf("%lld%lld", &L, &R);
    printf("%lld\n", count_classy(R) - count_classy(L - 1));
  }
}