#include <bits/stdc++.h>
using namespace std;
const int X = 30;
long long L, R, memo[X][4][2];
int T, len;
string s;
long long dfs(int cur, int M, int f) {
  if (M > 3) return 0;
  if (cur == len) return 1;
  if (memo[cur][M][f] != -1) return memo[cur][M][f];
  int v;
  if (f)
    v = s[cur] - '0';
  else
    v = 9;
  long long ans = 0;
  for (int i = 0; i <= v; i++) {
    if (i == 0 && f == 1 && i == v)
      ans += dfs(cur + 1, M, 1);
    else if (i == 0)
      ans += dfs(cur + 1, M, 0);
    else if (f == 1 && i == v)
      ans += dfs(cur + 1, M + 1, 1);
    else
      ans += dfs(cur + 1, M + 1, 0);
  }
  memo[cur][M][f] = ans;
  return ans;
}
long long solve(long long x) {
  s = to_string(x);
  len = s.length();
  memset(memo, -1, sizeof(memo));
  return dfs(0, 0, 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  while (T-- > 0) {
    cin >> L >> R;
    cout << solve(R) - solve(L - 1) << "\n";
  }
}

### Prompt

Your task is to create 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;
const int X = 30;
long long L, R, memo[X][4][2];
int T, len;
string s;
long long dfs(int cur, int M, int f) {
  if (M > 3) return 0;
  if (cur == len) return 1;
  if (memo[cur][M][f] != -1) return memo[cur][M][f];
  int v;
  if (f)
    v = s[cur] - '0';
  else
    v = 9;
  long long ans = 0;
  for (int i = 0; i <= v; i++) {
    if (i == 0 && f == 1 && i == v)
      ans += dfs(cur + 1, M, 1);
    else if (i == 0)
      ans += dfs(cur + 1, M, 0);
    else if (f == 1 && i == v)
      ans += dfs(cur + 1, M + 1, 1);
    else
      ans += dfs(cur + 1, M + 1, 0);
  }
  memo[cur][M][f] = ans;
  return ans;
}
long long solve(long long x) {
  s = to_string(x);
  len = s.length();
  memset(memo, -1, sizeof(memo));
  return dfs(0, 0, 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  while (T-- > 0) {
    cin >> L >> R;
    cout << solve(R) - solve(L - 1) << "\n";
  }
}

```