#include <bits/stdc++.h>
using namespace std;
long long dp[21][21][2];
long long digit[21];
long long solve(long long index, long long count, long long flag) {
  if (dp[index][count][flag] != -1) {
    return dp[index][count][flag];
  }
  if (index == 20) {
    if (count <= 3) {
      return 1;
    } else {
      return 0;
    }
  }
  long long res = 0;
  if (flag == 0) {
    long long di = digit[index];
    for (long long i = 0; i < di; i++) {
      res += solve(index + 1, count + (i != 0), 1);
    }
    res += solve(index + 1, count + (di != 0), 0);
  } else {
    for (long long i = 0; i < 10; i++) {
      res += solve(index + 1, count + (i != 0), 1);
    }
  }
  return dp[index][count][flag] = res;
}
long long count(long long num) {
  memset(dp, -1, sizeof(dp));
  memset(digit, 0, sizeof(digit));
  long long i = 19;
  while (num != 0) {
    long long d = (num % 10);
    digit[i--] = d;
    num /= 10;
  }
  return solve(0, 0, 0);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    cout << count(r) - count(l - 1) << endl;
  }
}

### Prompt

Please formulate 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;
long long dp[21][21][2];
long long digit[21];
long long solve(long long index, long long count, long long flag) {
  if (dp[index][count][flag] != -1) {
    return dp[index][count][flag];
  }
  if (index == 20) {
    if (count <= 3) {
      return 1;
    } else {
      return 0;
    }
  }
  long long res = 0;
  if (flag == 0) {
    long long di = digit[index];
    for (long long i = 0; i < di; i++) {
      res += solve(index + 1, count + (i != 0), 1);
    }
    res += solve(index + 1, count + (di != 0), 0);
  } else {
    for (long long i = 0; i < 10; i++) {
      res += solve(index + 1, count + (i != 0), 1);
    }
  }
  return dp[index][count][flag] = res;
}
long long count(long long num) {
  memset(dp, -1, sizeof(dp));
  memset(digit, 0, sizeof(digit));
  long long i = 19;
  while (num != 0) {
    long long d = (num % 10);
    digit[i--] = d;
    num /= 10;
  }
  return solve(0, 0, 0);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    cout << count(r) - count(l - 1) << endl;
  }
}

```