#include <bits/stdc++.h>
using namespace std;
const int N = 500 + 10;
int num[20];
int dp[20][4];
int dfs(int cur, int rest, bool flag) {
  if (cur == -1) {
    return 1;
  }
  if (!rest) return 1;
  if (!flag && ~dp[cur][rest]) return dp[cur][rest];
  int res = 0;
  int up = flag ? num[cur] : 9;
  for (int i = 0; i <= up; i++) {
    res += dfs(cur - 1, rest - bool(i), flag && i == up);
  }
  return flag ? res : dp[cur][rest] = res;
}
int solve(long long n) {
  int cnt = 0;
  while (n) {
    num[cnt++] = n % 10;
    n /= 10;
  }
  return dfs(cnt - 1, 3, true);
}
int main() {
  int t;
  cin >> t;
  long long l, r;
  memset(dp, -1, sizeof(dp));
  while (t--) {
    cin >> l >> r;
    cout << solve(r) - solve(l - 1) << endl;
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const int N = 500 + 10;
int num[20];
int dp[20][4];
int dfs(int cur, int rest, bool flag) {
  if (cur == -1) {
    return 1;
  }
  if (!rest) return 1;
  if (!flag && ~dp[cur][rest]) return dp[cur][rest];
  int res = 0;
  int up = flag ? num[cur] : 9;
  for (int i = 0; i <= up; i++) {
    res += dfs(cur - 1, rest - bool(i), flag && i == up);
  }
  return flag ? res : dp[cur][rest] = res;
}
int solve(long long n) {
  int cnt = 0;
  while (n) {
    num[cnt++] = n % 10;
    n /= 10;
  }
  return dfs(cnt - 1, 3, true);
}
int main() {
  int t;
  cin >> t;
  long long l, r;
  memset(dp, -1, sizeof(dp));
  while (t--) {
    cin >> l >> r;
    cout << solve(r) - solve(l - 1) << endl;
  }
  return 0;
}

```