#include <bits/stdc++.h>
using namespace std;
long long dp[19][4][2];
long long solve(long long pos, long long cnt, bool tight, string first) {
  if (pos == first.size()) {
    return 1;
  }
  if (dp[pos][cnt][tight] != -1) {
    return dp[pos][cnt][tight];
  }
  long long ret = 0;
  long long en;
  if (tight == 1) {
    en = first[pos] - '0';
  } else {
    en = 9;
  }
  for (int i = 0; i <= en; ++i) {
    if (i == 0) {
      ret += solve(pos + 1, cnt, tight & (i == en), first);
    } else if (i != 0 && cnt + 1 <= 3) {
      ret += solve(pos + 1, cnt + 1, tight & (i == en), first);
    }
  }
  return dp[pos][cnt][tight] = ret;
}
int main() {
  long long t;
  cin >> t;
  for (long long i = 0; i < t; ++i) {
    long long l, r;
    cin >> l >> r;
    memset(dp, -1, sizeof dp);
    string first = to_string(r);
    long long ans1 = solve(0, 0, 1, first);
    memset(dp, -1, sizeof dp);
    --l;
    first = to_string(l);
    long long ans2 = solve(0, 0, 1, first);
    cout << ans1 - ans2 << endl;
  }
}

### Prompt

Generate 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 namespace std;
long long dp[19][4][2];
long long solve(long long pos, long long cnt, bool tight, string first) {
  if (pos == first.size()) {
    return 1;
  }
  if (dp[pos][cnt][tight] != -1) {
    return dp[pos][cnt][tight];
  }
  long long ret = 0;
  long long en;
  if (tight == 1) {
    en = first[pos] - '0';
  } else {
    en = 9;
  }
  for (int i = 0; i <= en; ++i) {
    if (i == 0) {
      ret += solve(pos + 1, cnt, tight & (i == en), first);
    } else if (i != 0 && cnt + 1 <= 3) {
      ret += solve(pos + 1, cnt + 1, tight & (i == en), first);
    }
  }
  return dp[pos][cnt][tight] = ret;
}
int main() {
  long long t;
  cin >> t;
  for (long long i = 0; i < t; ++i) {
    long long l, r;
    cin >> l >> r;
    memset(dp, -1, sizeof dp);
    string first = to_string(r);
    long long ans1 = solve(0, 0, 1, first);
    memset(dp, -1, sizeof dp);
    --l;
    first = to_string(l);
    long long ans2 = solve(0, 0, 1, first);
    cout << ans1 - ans2 << endl;
  }
}

```