#include <bits/stdc++.h>
using namespace std;
vector<long long> inp;
long long lim;
long long dp[25][2][2][5];
long long DP(long long pos, long long isSmall, long long isStart,
             long long val) {
  if (val > 3) return 0;
  if (pos == lim) {
    return 1;
  }
  if (dp[pos][isSmall][isStart][val] != -1)
    return dp[pos][isSmall][isStart][val];
  long long ses = isSmall ? 9 : inp[pos];
  long long ret = 0;
  if (!isStart) {
    for (int i = 0; i <= ses; i++) {
      ret += DP(pos + 1, isSmall | i < inp[pos], 0, (i != 0) + val);
    }
  } else {
    for (int i = 1; i <= ses; i++) {
      ret += DP(pos + 1, isSmall | i < inp[pos], 0, (i != 0) + val);
    }
    ret += DP(pos + 1, 1, 1, 0);
  }
  return dp[pos][isSmall][isStart][val] = ret;
}
long long cal(long long x) {
  inp.clear();
  while (x) {
    inp.push_back(x % 10);
    x = x / 10;
  }
  reverse(inp.begin(), inp.end());
  lim = inp.size();
  memset(dp, -1, sizeof(dp));
  return DP(0, 0, 1, 0);
}
int main() {
  long long T;
  cin >> T;
  while (T--) {
    long long x, y;
    cin >> x >> y;
    cout << cal(y) - cal(x - 1) << endl;
  }
}

### 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;
vector<long long> inp;
long long lim;
long long dp[25][2][2][5];
long long DP(long long pos, long long isSmall, long long isStart,
             long long val) {
  if (val > 3) return 0;
  if (pos == lim) {
    return 1;
  }
  if (dp[pos][isSmall][isStart][val] != -1)
    return dp[pos][isSmall][isStart][val];
  long long ses = isSmall ? 9 : inp[pos];
  long long ret = 0;
  if (!isStart) {
    for (int i = 0; i <= ses; i++) {
      ret += DP(pos + 1, isSmall | i < inp[pos], 0, (i != 0) + val);
    }
  } else {
    for (int i = 1; i <= ses; i++) {
      ret += DP(pos + 1, isSmall | i < inp[pos], 0, (i != 0) + val);
    }
    ret += DP(pos + 1, 1, 1, 0);
  }
  return dp[pos][isSmall][isStart][val] = ret;
}
long long cal(long long x) {
  inp.clear();
  while (x) {
    inp.push_back(x % 10);
    x = x / 10;
  }
  reverse(inp.begin(), inp.end());
  lim = inp.size();
  memset(dp, -1, sizeof(dp));
  return DP(0, 0, 1, 0);
}
int main() {
  long long T;
  cin >> T;
  while (T--) {
    long long x, y;
    cin >> x >> y;
    cout << cal(y) - cal(x - 1) << endl;
  }
}

```