#include <bits/stdc++.h>
using namespace std;
int dig[22];
long long dp[22][4];
long long dfs(int len, int cnt, bool limit) {
  if (cnt > 3) return 0;
  if (!len) return cnt <= 3;
  long long& v = dp[len][cnt];
  if (!limit && v != -1) return v;
  int end = limit ? dig[len] : 9;
  int ans = 0;
  for (int i = 0; i < end + 1; ++i) {
    ans += dfs(len - 1, cnt + (i != 0), limit && i == end);
  }
  if (!limit) v = ans;
  return ans;
}
long long solve(long long x) {
  int idx = 0;
  memset(dp, -1, sizeof(dp));
  ;
  while (x) {
    dig[++idx] = x % 10;
    x /= 10;
  }
  return dfs(idx, 0, 1);
}
int main(int argc, char const* argv[]) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    cout << solve(r) - solve(l - 1) << endl;
  }
  return 0;
}

### 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;
int dig[22];
long long dp[22][4];
long long dfs(int len, int cnt, bool limit) {
  if (cnt > 3) return 0;
  if (!len) return cnt <= 3;
  long long& v = dp[len][cnt];
  if (!limit && v != -1) return v;
  int end = limit ? dig[len] : 9;
  int ans = 0;
  for (int i = 0; i < end + 1; ++i) {
    ans += dfs(len - 1, cnt + (i != 0), limit && i == end);
  }
  if (!limit) v = ans;
  return ans;
}
long long solve(long long x) {
  int idx = 0;
  memset(dp, -1, sizeof(dp));
  ;
  while (x) {
    dig[++idx] = x % 10;
    x /= 10;
  }
  return dfs(idx, 0, 1);
}
int main(int argc, char const* argv[]) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    cout << solve(r) - solve(l - 1) << endl;
  }
  return 0;
}

```