#include <bits/stdc++.h>
using namespace std;
long long T, dp[50][3][50];
vector<long long> v(20);
long long func(long long pos, long long flag, long long cnt) {
  if (pos == -1) {
    if (cnt <= 3) return 1;
    return 0;
  }
  if (dp[pos][flag][cnt] != -1) return dp[pos][flag][cnt];
  long long mx = (flag == 1) ? (v[pos]) : (9);
  long long ans = 0;
  for (long long i = 0; i < mx + 1; ++i) {
    if (i == 0)
      ans += func(pos - 1, ((i == mx) && flag), cnt);
    else
      ans += func(pos - 1, ((i == mx) && flag), cnt + 1);
  }
  return dp[pos][flag][cnt] = ans;
}
signed main() {
  cin >> T;
  for (long long t = 1; t < T + 1; ++t) {
    long long l, r, len = 0;
    cin >> l >> r;
    memset(dp, -1, sizeof dp);
    while (r > 0) {
      v[len] = r % 10;
      r = r / 10;
      len++;
    }
    long long ans = func(len - 1, 1, 0);
    l = l - 1;
    len = 0;
    while (l > 0) {
      v[len++] = l % 10;
      l = l / 10;
    }
    memset(dp, -1, sizeof dp);
    cout << ans - func(len - 1, 1, 0) << "\n";
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
long long T, dp[50][3][50];
vector<long long> v(20);
long long func(long long pos, long long flag, long long cnt) {
  if (pos == -1) {
    if (cnt <= 3) return 1;
    return 0;
  }
  if (dp[pos][flag][cnt] != -1) return dp[pos][flag][cnt];
  long long mx = (flag == 1) ? (v[pos]) : (9);
  long long ans = 0;
  for (long long i = 0; i < mx + 1; ++i) {
    if (i == 0)
      ans += func(pos - 1, ((i == mx) && flag), cnt);
    else
      ans += func(pos - 1, ((i == mx) && flag), cnt + 1);
  }
  return dp[pos][flag][cnt] = ans;
}
signed main() {
  cin >> T;
  for (long long t = 1; t < T + 1; ++t) {
    long long l, r, len = 0;
    cin >> l >> r;
    memset(dp, -1, sizeof dp);
    while (r > 0) {
      v[len] = r % 10;
      r = r / 10;
      len++;
    }
    long long ans = func(len - 1, 1, 0);
    l = l - 1;
    len = 0;
    while (l > 0) {
      v[len++] = l % 10;
      l = l / 10;
    }
    memset(dp, -1, sizeof dp);
    cout << ans - func(len - 1, 1, 0) << "\n";
  }
  return 0;
}

```