#include <bits/stdc++.h>
using namespace std;
int t, len = 20;
long long c[50][50], R[2];
long long solve(int pos, int ct, string &r) {
  long long ans = 0;
  if (pos > len - 1 || ct < 0) return 0;
  if (r[pos] - '0') {
    for (int k = 0; k <= min(ct - 1, len - 1 - pos); k++)
      ans += c[k][len - 1 - pos] * pow(9, k) * (r[pos] - '0' - 1);
    for (int k = 0; k <= min(ct, len - 1 - pos); k++)
      ans += c[k][len - 1 - pos] * pow(9, k);
  }
  ans += solve(pos + 1, (r[pos] == '0' ? ct : ct - 1), r);
  return ans;
}
int main() {
  c[0][0] = 1;
  for (int j = 1; j < 50; j++) {
    c[0][j] = 1;
    for (int i = 1; i <= j; i++) c[i][j] = c[i - 1][j - 1] + c[i][j - 1];
  }
  cin >> t;
  while (t--) {
    cin >> R[0] >> R[1];
    string s0(len - to_string(R[0]).length(), '0'),
        s1(len - to_string(R[1] + 1).length(), '0');
    printf("%I64d\n", solve(0, 3, s1.append(to_string(R[1] + 1))) -
                          solve(0, 3, s0.append(to_string(R[0]))));
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int t, len = 20;
long long c[50][50], R[2];
long long solve(int pos, int ct, string &r) {
  long long ans = 0;
  if (pos > len - 1 || ct < 0) return 0;
  if (r[pos] - '0') {
    for (int k = 0; k <= min(ct - 1, len - 1 - pos); k++)
      ans += c[k][len - 1 - pos] * pow(9, k) * (r[pos] - '0' - 1);
    for (int k = 0; k <= min(ct, len - 1 - pos); k++)
      ans += c[k][len - 1 - pos] * pow(9, k);
  }
  ans += solve(pos + 1, (r[pos] == '0' ? ct : ct - 1), r);
  return ans;
}
int main() {
  c[0][0] = 1;
  for (int j = 1; j < 50; j++) {
    c[0][j] = 1;
    for (int i = 1; i <= j; i++) c[i][j] = c[i - 1][j - 1] + c[i][j - 1];
  }
  cin >> t;
  while (t--) {
    cin >> R[0] >> R[1];
    string s0(len - to_string(R[0]).length(), '0'),
        s1(len - to_string(R[1] + 1).length(), '0');
    printf("%I64d\n", solve(0, 3, s1.append(to_string(R[1] + 1))) -
                          solve(0, 3, s0.append(to_string(R[0]))));
  }
  return 0;
}

```