1) Abridged problem statement
- Count the number of S-digit sequences (S between 1 and 20), digits 0–9, leading zeros allowed, such that the digits first go in non-decreasing order and then in non-increasing order (either part may be empty). Output the count.

2) Detailed editorial
- Model: A valid ticket is unimodal with a single “turn”: first non-decreasing, then non-increasing. Equal digits are allowed anywhere.
- Digit DP:
  - State dp[pos][phase][last]:
    - pos: how many digits have been placed so far (0..S).
    - phase: 0 = still in the non-decreasing (up) part, 1 = in the non-increasing (down) part.
    - last: the last digit placed (0..9).
  - Transitions:
    - If phase = 0:
      - Choose next digit d in [last..9] and stay in phase 0 (still non-decreasing).
      - Choose next digit d in [0..last-1] and switch to phase 1 (start the non-increasing part).
    - If phase = 1:
      - Choose next digit d in [0..last] (must be non-increasing).
  - Base case: If pos == S, return 1 (a complete sequence is formed).
  - Start: rec(0, 0, 0) because leading zeros are allowed; we begin in the up phase with virtual last digit 0. The first real digit d ≥ 0 is always valid, so this enumerates all sequences.
  - Complexity: O(S * 10 states * 10 transitions) ≈ 2000–4000 operations, trivial for S ≤ 20. Use 64-bit integers.
- Why no double counting: We switch to the down phase only when we actually decrease (pick a smaller digit); once switched, we never return to the up phase. Each unimodal sequence has exactly one such trajectory in this DP.
- Closed-form combinatorial formula (optional):
  - Let m be the maximum digit in the sequence (0..9). Choose counts of digits on the left (non-decreasing) and right (non-increasing) with the pivot being the last occurrence of the maximum digit m (so m appears only on the left). If a_m is the count of m on the left, we must have a_m ≥ 1. Let a'_m = a_m - 1 ≥ 0.
  - Counting nonnegative integer solutions:
    - Variables: a_0..a_{m-1}, a'_m, b_0..b_{m-1} (total 2m+1 variables).
    - Sum equals S - 1.
    - Number of solutions: C(S + 2m - 1, 2m).
  - Total answer: sum_{m=0}^{9} C(S + 2m - 1, 2m).
  - Checks: S=1 → sum_{m=0}^9 1 = 10; S=3 → 715 (matches sample).
- Either approach works; the DP is simplest to implement and is used in the provided solution.

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;
};

int S;

void read() { cin >> S; }

vector<vector<vector<int64_t>>> dp;

int64_t rec(int pos, int state, int last_digit) {
    if(pos == S) {
        return 1;
    }

    int64_t& memo = dp[pos][state][last_digit];
    if(memo != -1) {
        return memo;
    }

    memo = 0;

    if(state == 0) {
        for(int d = last_digit; d <= 9; d++) {
            memo += rec(pos + 1, 0, d);
        }
        for(int d = 0; d < last_digit; d++) {
            memo += rec(pos + 1, 1, d);
        }
    } else {
        for(int d = 0; d <= last_digit; d++) {
            memo += rec(pos + 1, 1, d);
        }
    }

    return memo;
}

void solve() {
    // This is a classical problem on digit DP. The state we want to keep is:
    //    dp[pos][state][last_digit].
    //        - pos is the position, starting from S - 1.
    //        - state = 0 means that we are building the increasing prefix,
    //          while state = 1 means that we are building the decreasing. We
    //          start from 0 state.
    //        - last_digit is the last digit we placed (0 at the beginning).

    dp.assign(S, vector<vector<int64_t>>(2, vector<int64_t>(10, -1)));
    cout << rec(0, 0, 0) << '\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 with detailed comments
```python
import sys
from functools import lru_cache

def main():
    data = sys.stdin.read().strip().split()
    S = int(data[0])  # Length of the ticket number (1 <= S <= 20)

    # Digit DP with memoization.
    # State:
    #   pos   : how many digits have been placed (0..S)
    #   phase : 0 = non-decreasing phase, 1 = non-increasing phase
    #   last  : value of the last placed digit (0..9)
    @lru_cache(maxsize=None)
    def rec(pos: int, phase: int, last: int) -> int:
        # Base case: completed S digits -> count as 1 valid ticket
        if pos == S:
            return 1

        total = 0
        if phase == 0:
            # Still non-decreasing:
            # - choose digit >= last and stay in phase 0
            for d in range(last, 10):
                total += rec(pos + 1, 0, d)
            # - choose digit < last and switch to phase 1
            for d in range(0, last):
                total += rec(pos + 1, 1, d)
        else:
            # Non-increasing phase: must pick digit <= last
            for d in range(0, last + 1):
                total += rec(pos + 1, 1, d)

        return total

    # Start at position 0, in non-decreasing phase, with last=0
    # (leading zeros allowed so this enumerates all).
    ans = rec(0, 0, 0)
    print(ans)

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

5) Compressed editorial
- Use digit DP with state (pos, phase, last). Start with (0, up, last=0).
- Transitions: in up-phase choose d ≥ last and stay up, or d < last to switch down; in down-phase choose d ≤ last.
- Base: pos == S -> 1. Complexity O(S * 10 * 10), fits easily.
- Optional closed form: sum_{m=0}^{9} C(S + 2m - 1, 2m).
