<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

502. Digits Permutation
Time limit per test: 0.25 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

Andrew has just made a breakthrough in the world of number 17: he realized that it's rather easy to permute the digits in almost any given number to get a number divisible by 17. You are given a positive integer n. You must find a permutation of its digits that is divisible by 17.
Input
Input file contains single integer n, 1 ≤ n ≤ 1017.
Output
Output any permutation of digits of n that is divisible by 17. The output permutation may not start with a zero. If there is no such permutation, output -1.
Example(s)
sample input
sample output
17
17

sample input
sample output
2242223
2222342

sample input
sample output
239
-1

<|response|>
1. Abridged Problem Statement
Given a positive integer n (up to 10^17 in value, i.e., up to 17 digits), find any permutation of its digits (not starting with '0') that is divisible by 17. If no such permutation exists, print −1.

2. Key Observations
- The number of digits is at most 17, so there are at most 2^17 ≃ 131 072 subsets of positions.
- Brute-forcing all permutations (17! ≃ 3.5·10^14) is impossible, but iterating over subsets with DP is feasible.
- We only care about divisibility by 17, i.e., the remainder modulo 17.
- We must forbid leading zeros: the first chosen digit cannot be '0'.

3. Full Solution Approach
We use a bitmask DP over subsets of digit-positions. Let s be the input string of digits, n = |s|. We index digits 0…n−1.

Define DP state:
  dp[mask][r] = the minimum numeric value (as a 64-bit integer) that can be formed by using exactly the set of positions in 'mask', and whose value mod 17 is r.
Here mask is an integer 0…(1<<n)−1, r is in 0…16.

Initialization:
  dp[0][0] = 0   (empty prefix, remainder 0)
  dp[0][r] = +∞ for r ≠ 0.

Transition:
  For each mask from 0 to (1<<n)−1:
    1. Count how many non-zero digits are already used in mask. If that count is zero, then we have not placed any non-zero digit yet, so appending a '0' next would make a leading zero—skip those transitions.
    2. For each position i not in mask:
         let d = s[i]−'0'.
         If d==0 and we have no non-zero yet, continue.
         newMask = mask | (1<<i).
         For each old remainder r in 0…16:
           if dp[mask][r] is infinite, skip.
           newRem = (r*10 + d) % 17
           newVal = dp[mask][r]*10 + d
           dp[newMask][newRem] = min(dp[newMask][newRem], newVal)

Answer:
  Look at dp[(1<<n)−1][0]. If it is still ∞, print −1; otherwise print that minimum value.

Time Complexity: O(2^n · n · 17), which for n≤17 is about 3·10^7 operations.

4. C++ Implementation with Detailed Comments
```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;
};

string s;

void read() { cin >> s; }

void solve() {
    // Bitmask DP over which digit positions have already been placed.
    //
    // - dp[mask][r] = the smallest number we can build by placing exactly
    //   the digits in mask, whose value modulo 17 equals r. We extend a
    //   state by appending one not-yet-used digit, keeping the minimal
    //   value so the final answer is the smallest valid permutation.
    //
    // - To avoid a leading zero we forbid placing a '0' as the very first
    //   digit (when no non-zero digit has been placed yet, i.e. the count
    //   of non-zero digits already in mask is 0).
    //
    // - The answer is dp[full][0]; if it stays at infinity no permutation
    //   is divisible by 17 and we print -1.

    if(s == "0") {
        cout << s << '\n';
        return;
    }

    int n = s.size();
    vector<vector<int64_t>> dp(
        1 << n, vector<int64_t>(17, numeric_limits<int64_t>::max())
    );
    dp[0][0] = 0;
    for(int mask = 0; mask < (1 << n); mask++) {
        int cnt_non_zero = 0;
        for(int i = 0; i < n; i++) {
            if((mask & (1 << i)) && s[i] != '0') {
                cnt_non_zero++;
            }
        }

        for(int i = 0; i < n; i++) {
            if((s[i] == '0' && cnt_non_zero == 0) || (mask & (1 << i))) {
                continue;
            }

            for(int j = 0; j < 17; j++) {
                if(dp[mask][j] == numeric_limits<int64_t>::max()) {
                    continue;
                }
                int nmask = mask | (1 << i);
                int nval = (j * 10 + s[i] - '0') % 17;
                dp[nmask][nval] =
                    min(dp[nmask][nval], dp[mask][j] * 10 + s[i] - '0');
            }
        }
    }

    int64_t ans = dp[(1 << n) - 1][0];
    if(ans == numeric_limits<int64_t>::max()) {
        cout << "-1\n";
        return;
    }

    cout << ans << '\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;
}
```

5. Python Implementation with Detailed Comments
```python
def find_permutation_div_by_17(s):
    """
    Given a string of digits s, find the minimum permutation
    (no leading zero) divisible by 17, or return '-1'.
    """
    n = len(s)
    # Special case
    if s == "0":
        return "0"

    FULL = 1 << n
    INF = 10**30
    # dp[mask][r] = minimum integer value formed, or INF if impossible
    dp = [ [INF]*17 for _ in range(FULL) ]
    dp[0][0] = 0

    # Precompute digits as ints
    digits = list(map(int, s))

    for mask in range(FULL):
        # Count how many non-zero digits used so far
        cnt_non_zero = 0
        for i in range(n):
            if (mask >> i) & 1 and digits[i] != 0:
                cnt_non_zero += 1

        # Try to add each unused index i next
        for i in range(n):
            if (mask >> i) & 1:
                continue
            d = digits[i]
            # Avoid leading zero if we have not yet placed a non-zero digit
            if d == 0 and cnt_non_zero == 0:
                continue

            new_mask = mask | (1 << i)
            # Go through each old remainder
            for r in range(17):
                old_val = dp[mask][r]
                if old_val == INF:
                    continue
                new_r = (r * 10 + d) % 17
                new_val = old_val * 10 + d
                # Take the minimum numeric value
                if new_val < dp[new_mask][new_r]:
                    dp[new_mask][new_r] = new_val

    ans = dp[FULL - 1][0]
    return str(ans) if ans < INF else "-1"

if __name__ == "__main__":
    s = input().strip()
    print(find_permutation_div_by_17(s))
```
