#include <bits/stdc++.h>
using namespace std;
long long dp[25][25][2];
vector<int> digits;
long long calc(int pos, int cnt, int f) {
  if (cnt > 3) return 0;
  if (pos == digits.size()) return 1;
  if (dp[pos][cnt][f] != -1) return dp[pos][cnt][f];
  int add;
  if (f)
    add = 9;
  else
    add = digits[pos];
  long long res = 0;
  for (int currdigit = 0; currdigit <= add; currdigit++) {
    int newf = f;
    int newcnt = cnt;
    if (currdigit) newcnt++;
    if (!f && currdigit < add) newf = 1;
    if (newcnt <= 3) res += calc(pos + 1, newcnt, newf);
  }
  return dp[pos][cnt][f] = res;
}
long long solve(long long num) {
  digits.clear();
  while (num) {
    digits.push_back(num % 10);
    num /= 10;
  }
  reverse(digits.begin(), digits.end());
  memset(dp, -1, sizeof(dp));
  long long res = calc(0, 0, 0);
  return res;
}
int main() {
  int q;
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    cout << solve(r) - solve(l - 1) << "\n";
  }
}

### 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;
long long dp[25][25][2];
vector<int> digits;
long long calc(int pos, int cnt, int f) {
  if (cnt > 3) return 0;
  if (pos == digits.size()) return 1;
  if (dp[pos][cnt][f] != -1) return dp[pos][cnt][f];
  int add;
  if (f)
    add = 9;
  else
    add = digits[pos];
  long long res = 0;
  for (int currdigit = 0; currdigit <= add; currdigit++) {
    int newf = f;
    int newcnt = cnt;
    if (currdigit) newcnt++;
    if (!f && currdigit < add) newf = 1;
    if (newcnt <= 3) res += calc(pos + 1, newcnt, newf);
  }
  return dp[pos][cnt][f] = res;
}
long long solve(long long num) {
  digits.clear();
  while (num) {
    digits.push_back(num % 10);
    num /= 10;
  }
  reverse(digits.begin(), digits.end());
  memset(dp, -1, sizeof(dp));
  long long res = calc(0, 0, 0);
  return res;
}
int main() {
  int q;
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    cout << solve(r) - solve(l - 1) << "\n";
  }
}

```