#include <bits/stdc++.h>
using namespace std;
const int N = 20;
long long dp[N][2][4];
vector<int> num(N);
long long cn(int cur, bool f, int k) {
  if (k > 3) return 0;
  if (cur == N) return (k > 0);
  if (dp[cur][f][k] == -1) {
    int limit = 9;
    if (f) limit = num[cur];
    long long res = 0;
    for (int i = 0; i <= limit; i++) {
      if (f && i == limit)
        res += cn(cur + 1, true, k + (i > 0));
      else
        res += cn(cur + 1, false, k + (i > 0));
    }
    dp[cur][f][k] = res;
  }
  return dp[cur][f][k];
}
long long solve(long long n) {
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < N; i++) num[i] = 0;
  for (int i = N - 1; i >= 0 && n > 0; i--, n /= 10) num[i] = n % 10;
  return cn(0, true, 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t-- > 0) {
    long long l, r;
    cin >> l >> r;
    cout << solve(r) - solve(l - 1) << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 N = 20;
long long dp[N][2][4];
vector<int> num(N);
long long cn(int cur, bool f, int k) {
  if (k > 3) return 0;
  if (cur == N) return (k > 0);
  if (dp[cur][f][k] == -1) {
    int limit = 9;
    if (f) limit = num[cur];
    long long res = 0;
    for (int i = 0; i <= limit; i++) {
      if (f && i == limit)
        res += cn(cur + 1, true, k + (i > 0));
      else
        res += cn(cur + 1, false, k + (i > 0));
    }
    dp[cur][f][k] = res;
  }
  return dp[cur][f][k];
}
long long solve(long long n) {
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < N; i++) num[i] = 0;
  for (int i = N - 1; i >= 0 && n > 0; i--, n /= 10) num[i] = n % 10;
  return cn(0, true, 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t-- > 0) {
    long long l, r;
    cin >> l >> r;
    cout << solve(r) - solve(l - 1) << "\n";
  }
  return 0;
}

```