#include <bits/stdc++.h>
using namespace std;
int a[25];
long long dp[25][25];
long long dfs(int len, int sum, bool limit) {
  if (len == -1) return sum < 4;
  if (!limit && dp[len][sum] != -1) return dp[len][sum];
  long long ans = 0;
  int up = limit ? a[len] : 9;
  for (int i = 0; i <= up; i++) {
    ans += dfs(len - 1, sum + (i != 0), i == a[len] && limit);
  }
  if (!limit) dp[len][sum] = ans;
  return ans;
}
long long fun(long long x) {
  int len = 0;
  while (x) {
    a[len++] = x % 10;
    x /= 10;
  }
  return dfs(len - 1, 0, 1);
}
int main() {
  long long n, m, t;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    memset(dp, -1, sizeof(dp));
    cout << fun(m) - fun(n - 1) << endl;
  }
  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 a[25];
long long dp[25][25];
long long dfs(int len, int sum, bool limit) {
  if (len == -1) return sum < 4;
  if (!limit && dp[len][sum] != -1) return dp[len][sum];
  long long ans = 0;
  int up = limit ? a[len] : 9;
  for (int i = 0; i <= up; i++) {
    ans += dfs(len - 1, sum + (i != 0), i == a[len] && limit);
  }
  if (!limit) dp[len][sum] = ans;
  return ans;
}
long long fun(long long x) {
  int len = 0;
  while (x) {
    a[len++] = x % 10;
    x /= 10;
  }
  return dfs(len - 1, 0, 1);
}
int main() {
  long long n, m, t;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    memset(dp, -1, sizeof(dp));
    cout << fun(m) - fun(n - 1) << endl;
  }
  return 0;
}

```