#include <bits/stdc++.h>
using namespace std;
int digit[20];
bool is[20];
long long mods(int bit, long long mai_mic, long long egal) {
  if (bit < 0) return mai_mic + egal;
  if (!is[bit]) {
    if (!digit[bit]) return mods(bit - 1, mai_mic, egal);
    return mods(bit - 1, mai_mic + egal, 0);
  }
  if (digit[bit] >= 2)
    return mods(bit - 1, mai_mic * 9 + egal * (digit[bit] - 1), egal);
  else if (digit[bit] == 0)
    return mods(bit - 1, mai_mic * 9, 0);
  return mods(bit - 1, mai_mic * 9, egal);
}
long long mod(long long x) {
  int bmax = 0;
  for (int i = 0; x; i++, bmax++) digit[i] = x % 10, x /= 10;
  long long ans = 0;
  for (int i = 0; i < bmax; i++) {
    for (int j = i; j < bmax; j++) {
      for (int k = j; k < bmax; k++) {
        if (i == j && k != j) continue;
        is[i] = is[j] = is[k] = 1;
        ans += mods(bmax - 1, 0, 1);
        is[i] = is[j] = is[k] = 0;
      }
    }
  }
  return ans;
}
int main() {
  int n;
  long long a, b, k;
  cin >> n;
  while (n--) {
    cin >> a >> b;
    cout << mod(b) - mod(a - 1) << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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 digit[20];
bool is[20];
long long mods(int bit, long long mai_mic, long long egal) {
  if (bit < 0) return mai_mic + egal;
  if (!is[bit]) {
    if (!digit[bit]) return mods(bit - 1, mai_mic, egal);
    return mods(bit - 1, mai_mic + egal, 0);
  }
  if (digit[bit] >= 2)
    return mods(bit - 1, mai_mic * 9 + egal * (digit[bit] - 1), egal);
  else if (digit[bit] == 0)
    return mods(bit - 1, mai_mic * 9, 0);
  return mods(bit - 1, mai_mic * 9, egal);
}
long long mod(long long x) {
  int bmax = 0;
  for (int i = 0; x; i++, bmax++) digit[i] = x % 10, x /= 10;
  long long ans = 0;
  for (int i = 0; i < bmax; i++) {
    for (int j = i; j < bmax; j++) {
      for (int k = j; k < bmax; k++) {
        if (i == j && k != j) continue;
        is[i] = is[j] = is[k] = 1;
        ans += mods(bmax - 1, 0, 1);
        is[i] = is[j] = is[k] = 0;
      }
    }
  }
  return ans;
}
int main() {
  int n;
  long long a, b, k;
  cin >> n;
  while (n--) {
    cin >> a >> b;
    cout << mod(b) - mod(a - 1) << '\n';
  }
  return 0;
}

```