#include <bits/stdc++.h>
using std::cin;
using std::cout;
const int maxn = 2e5 + 3, INF = 0x3f3f3f3f;
typedef int arr[maxn];
int c[19];
long long f[19][4];
long long Dfs(int x, int y, bool flag) {
  if (y > 3) return 0;
  if (!x) return 1;
  if (!flag && ~f[x][y]) return f[x][y];
  int lim = flag ? c[x] : 9;
  long long ans = 0;
  for (int i = (0); i <= (lim); ++i) {
    ans += Dfs(x - 1, y + (bool)i, flag && i == lim);
  }
  return flag ? ans : f[x][y] = ans;
}
long long Solve(long long x) {
  int tot = 0;
  while (x) {
    c[++tot] = x % 10;
    x /= 10;
  }
  return Dfs(tot, 0, 1);
}
int main() {
  int n;
  cin >> n;
  memset(f, -1, sizeof f);
  while (n--) {
    long long l, r;
    cin >> l >> r;
    cout << Solve(r) - Solve(l - 1) << '\n';
  }
  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 std::cin;
using std::cout;
const int maxn = 2e5 + 3, INF = 0x3f3f3f3f;
typedef int arr[maxn];
int c[19];
long long f[19][4];
long long Dfs(int x, int y, bool flag) {
  if (y > 3) return 0;
  if (!x) return 1;
  if (!flag && ~f[x][y]) return f[x][y];
  int lim = flag ? c[x] : 9;
  long long ans = 0;
  for (int i = (0); i <= (lim); ++i) {
    ans += Dfs(x - 1, y + (bool)i, flag && i == lim);
  }
  return flag ? ans : f[x][y] = ans;
}
long long Solve(long long x) {
  int tot = 0;
  while (x) {
    c[++tot] = x % 10;
    x /= 10;
  }
  return Dfs(tot, 0, 1);
}
int main() {
  int n;
  cin >> n;
  memset(f, -1, sizeof f);
  while (n--) {
    long long l, r;
    cin >> l >> r;
    cout << Solve(r) - Solve(l - 1) << '\n';
  }
  return 0;
}

```