#include <bits/stdc++.h>
using namespace std;
long long C[20][20];
long long cal(long long dic, long long cur) {
  cur = min(dic, cur);
  long long ans = 0;
  long long ppow = 1;
  for (int i = 0; i <= cur; ++i) {
    ans += C[dic][i] * ppow;
    ppow *= 9;
  }
  return ans;
}
long long Solve(long long num, int cur) {
  if (cur == 0 || num == 0) return (long long)1;
  int dic = 1;
  long long tnum = num;
  while (tnum /= 10) ++dic;
  long long tmp = 1;
  for (int i = 2; i <= dic; ++i) tmp *= 10;
  long long k = num / tmp;
  long long ans = (k - 1) * cal(dic - 1, cur - 1);
  ans += cal(dic - 1, cur);
  ans += Solve(num % tmp, cur - 1);
  return ans;
}
int main() {
  C[0][0] = 1;
  for (int i = 1; i <= 18; ++i) {
    for (int j = 0; j <= i; ++j) {
      C[i][j] = C[i - 1][j];
      if (j != 0) C[i][j] += C[i - 1][j - 1];
    }
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    long long L, R;
    scanf("%lld %lld", &L, &R);
    printf("%lld\n", Solve(R, 3) - Solve(L - 1, 3));
  }
}

### Prompt

Create a solution in CPP for the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long C[20][20];
long long cal(long long dic, long long cur) {
  cur = min(dic, cur);
  long long ans = 0;
  long long ppow = 1;
  for (int i = 0; i <= cur; ++i) {
    ans += C[dic][i] * ppow;
    ppow *= 9;
  }
  return ans;
}
long long Solve(long long num, int cur) {
  if (cur == 0 || num == 0) return (long long)1;
  int dic = 1;
  long long tnum = num;
  while (tnum /= 10) ++dic;
  long long tmp = 1;
  for (int i = 2; i <= dic; ++i) tmp *= 10;
  long long k = num / tmp;
  long long ans = (k - 1) * cal(dic - 1, cur - 1);
  ans += cal(dic - 1, cur);
  ans += Solve(num % tmp, cur - 1);
  return ans;
}
int main() {
  C[0][0] = 1;
  for (int i = 1; i <= 18; ++i) {
    for (int j = 0; j <= i; ++j) {
      C[i][j] = C[i - 1][j];
      if (j != 0) C[i][j] += C[i - 1][j - 1];
    }
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    long long L, R;
    scanf("%lld %lld", &L, &R);
    printf("%lld\n", Solve(R, 3) - Solve(L - 1, 3));
  }
}

```