#include <bits/stdc++.h>
using namespace std;
const long long N = 22;
long long a[N];
long long dp[N][N][N];
long long DP(long long pos, long long sum, long long limit) {
  if (!pos) return 1;
  if (!limit && dp[pos][sum][limit] != -1) return dp[pos][sum][limit];
  long long high = limit ? a[pos] : 9;
  long long ans = 0;
  for (long long i = 0; i <= high; i++) {
    if (sum == 3 && i != 0) continue;
    ans += DP(pos - 1, sum + (i != 0), limit && (i == high));
  }
  if (!limit) dp[pos][sum][limit] = ans;
  return ans;
}
long long slove(long long n) {
  long long len = 0;
  while (n) {
    a[++len] = n % 10;
    n = n / 10;
  }
  return DP(len, 0, 1);
}
signed main() {
  memset(dp, -1, sizeof(dp));
  long long T;
  cin >> T;
  while (T--) {
    long long L, R;
    cin >> L >> R;
    cout << slove(R) - slove(L - 1) << '\n';
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long N = 22;
long long a[N];
long long dp[N][N][N];
long long DP(long long pos, long long sum, long long limit) {
  if (!pos) return 1;
  if (!limit && dp[pos][sum][limit] != -1) return dp[pos][sum][limit];
  long long high = limit ? a[pos] : 9;
  long long ans = 0;
  for (long long i = 0; i <= high; i++) {
    if (sum == 3 && i != 0) continue;
    ans += DP(pos - 1, sum + (i != 0), limit && (i == high));
  }
  if (!limit) dp[pos][sum][limit] = ans;
  return ans;
}
long long slove(long long n) {
  long long len = 0;
  while (n) {
    a[++len] = n % 10;
    n = n / 10;
  }
  return DP(len, 0, 1);
}
signed main() {
  memset(dp, -1, sizeof(dp));
  long long T;
  cin >> T;
  while (T--) {
    long long L, R;
    cin >> L >> R;
    cout << slove(R) - slove(L - 1) << '\n';
  }
  return 0;
}

```