1. Abridged Problem Statement
Given a positive integer N (1 ≤ N ≤ 2³¹–1). Define aₖ as the decimal number obtained by concatenating the integers 1,2,…,k (for example, a₄=1234). Count how many of the terms a₁, a₂, …, a_N are divisible by 3.

2. Detailed Editorial

  a) Key observation—divisibility by 3 via digit sum:
     A number is divisible by 3 iff the sum of its digits is divisible by 3.

  b) Relate aₖ’s digit sum to 1+2+…+k:
     When you concatenate 1,2,…,k, the total digit-sum is the sum of the digit-sums of each integer i from 1 to k. Furthermore, each i ≡ (sum of its digits) mod 3. Therefore
       aₖ mod 3  = (sum of all digits of aₖ) mod 3
                  = (∑_{i=1}^k (digitsum(i))) mod 3
                  = (∑_{i=1}^k i) mod 3
                  = [k·(k+1)/2] mod 3.

  c) Solve k(k+1)/2 ≡ 0 (mod 3).
     Since 2 is invertible mod 3, this is equivalent to k·(k+1) ≡ 0 (mod 3).
     A product modulo 3 is zero if and only if one factor is 0 mod 3. Hence:
       k ≡ 0 (mod 3)  or  k+1 ≡ 0 (mod 3)  ⇔  k ≡ 0 or 2 (mod 3).

  d) Counting in [1..N]:
     In each block of 3 consecutive k’s, exactly two satisfy (k mod 3 ∈ {0,2}).
     Let m = ⌊N/3⌋. Then from the full blocks we get 2m counts. For the remainder r = N mod 3:
       • r = 0 → no extra
       • r = 1 → k=3m+1 (mod 3=1) → no extra
       • r = 2 → k=3m+1(mod 3=1) + k=3m+2(mod 3=2) → one extra
     Thus answer = 2·⌊N/3⌋ + [r == 2 ? 1 : 0].

  e) Complexity and limits:
     All operations are O(1), and N fits in 64 bits.

3. C++ Solution
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int64_t n;

void read() { cin >> n; }

void solve() {
    // The k-th term 123...k is divisible by 3 exactly when its digit sum, which
    // is 1+2+...+k = k(k+1)/2, is divisible by 3, i.e. when k mod 3 is 0 or 2.
    // So among k = 1..n there are 2 such terms per full block of 3, plus one
    // extra when n mod 3 == 2.

    int64_t answer = n / 3 * 2;
    if(n % 3 == 2) {
        answer++;
    }

    cout << answer << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution
```python
import sys

def main():
    # Read N from standard input
    data = sys.stdin.read().strip()
    n = int(data)

    # Each full group of 3 numbers (k = 1..3, 4..6, ...) has exactly two k's with
    # (k mod 3 == 0 or 2). So full_groups = n // 3, each gives 2 counts.
    full_groups = n // 3
    answer = full_groups * 2

    # For the leftover 1 or 2 values:
    # - If leftover == 2, that includes one k with k mod 3 == 2
    # - Otherwise no extra
    if n % 3 == 2:
        answer += 1

    # Print the result
    print(answer)

if __name__ == "__main__":
    main()
```

5. Compressed Editorial
Every concatenated term aₖ has digit-sum ≡ 1+2+…+k mod 3, i.e. k(k+1)/2 mod 3.
k(k+1)/2 ≡ 0 (mod 3) ⇔ k(k+1) ≡ 0 (mod 3) ⇔ k ≡ 0 or 2 (mod 3).
Out of each 3 consecutive k’s, two qualify. Answer = 2·⌊N/3⌋ + (N mod 3 == 2 ? 1 : 0).
