<|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;

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

    string s;
    cin >> s;
    // Special case: "0" is itself divisible by 17
    if (s == "0") {
        cout << s << "\n";
        return 0;
    }

    int n = s.size();
    int FULL = 1 << n;
    const long long INF = LLONG_MAX;

    // dp[mask][r]: minimum value formed by selecting positions in 'mask'
    //              with remainder r modulo 17.
    vector< array<long long, 17> > dp(FULL);
    // Initialize all states to INF
    for (int mask = 0; mask < FULL; mask++) {
        for (int r = 0; r < 17; r++) {
            dp[mask][r] = INF;
        }
    }
    dp[0][0] = 0;  // empty prefix, remainder 0

    // Pre-convert digits
    vector<int> digit(n);
    for (int i = 0; i < n; i++) {
        digit[i] = s[i] - '0';
    }

    // Iterate over every subset of positions
    for (int mask = 0; mask < FULL; mask++) {
        // Count non-zero digits already used to prevent leading zero
        int cntNonZero = 0;
        for (int i = 0; i < n; i++) {
            if ((mask & (1 << i)) && digit[i] != 0) {
                cntNonZero++;
            }
        }

        // Try to append each unused position i
        for (int i = 0; i < n; i++) {
            if (mask & (1 << i)) continue;       // already used
            if (digit[i] == 0 && cntNonZero == 0) 
                continue;  // would lead with zero

            int newMask = mask | (1 << i);
            int d = digit[i];

            // Transition over all old remainders
            for (int r = 0; r < 17; r++) {
                long long oldVal = dp[mask][r];
                if (oldVal == INF) continue;

                int newRem = (r * 10 + d) % 17;
                long long newVal = oldVal * 10 + d;
                if (newVal < dp[newMask][newRem]) {
                    dp[newMask][newRem] = newVal;
                }
            }
        }
    }

    long long answer = dp[FULL - 1][0];
    if (answer == INF) {
        cout << "-1\n";
    } else {
        cout << answer << "\n";
    }
    return 0;
}
```

5. Python Implementation with Detailed Comments  
```python
def find_div_by_17(s):
    # Special-case: "0"
    if s == "0":
        return "0"

    n = len(s)
    FULL = 1 << n
    INF = 10**30

    # Convert to list of ints
    digits = list(map(int, s))

    # dp[mask][r] = minimal integer value with positions=mask, remainder=r
    dp = [ [INF]*17 for _ in range(FULL) ]
    dp[0][0] = 0

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

        # Try to append each unused digit
        for i in range(n):
            if (mask >> i) & 1:
                continue  # already used
            if digits[i] == 0 and cnt_non_zero == 0:
                continue  # leading zero not allowed

            new_mask = mask | (1 << i)
            d = digits[i]

            # For 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
                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_div_by_17(s))
```