#include <bits/stdc++.h>
using namespace std;
long long dp[20][5][2];
string s;
long long l, r;
long long solve(long long pos, long long cnt, bool tight) {
  if (pos == s.size()) return 1;
  if (dp[pos][cnt][tight] != -1) return dp[pos][cnt][tight];
  long long ret = 0;
  long long en = tight ? (s[pos] - '0') : 9;
  for (long long i = 0; i <= en; i++) {
    long long c = cnt + (i > 0);
    if (c <= 3) {
      ret += solve(pos + 1, c, tight & (i == en));
    }
  }
  return dp[pos][cnt][tight] = ret;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> l >> r;
    s = to_string(r);
    memset(dp, -1, sizeof dp);
    long long ans = solve(0, 0, 1);
    l -= 1;
    s = to_string(l);
    memset(dp, -1, sizeof dp);
    ans -= solve(0, 0, 1);
    cout << ans << endl;
  }
  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 dp[20][5][2];
string s;
long long l, r;
long long solve(long long pos, long long cnt, bool tight) {
  if (pos == s.size()) return 1;
  if (dp[pos][cnt][tight] != -1) return dp[pos][cnt][tight];
  long long ret = 0;
  long long en = tight ? (s[pos] - '0') : 9;
  for (long long i = 0; i <= en; i++) {
    long long c = cnt + (i > 0);
    if (c <= 3) {
      ret += solve(pos + 1, c, tight & (i == en));
    }
  }
  return dp[pos][cnt][tight] = ret;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> l >> r;
    s = to_string(r);
    memset(dp, -1, sizeof dp);
    long long ans = solve(0, 0, 1);
    l -= 1;
    s = to_string(l);
    memset(dp, -1, sizeof dp);
    ans -= solve(0, 0, 1);
    cout << ans << endl;
  }
  return 0;
}

```