<|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.

205. Quantization Problem
time limit per test: 0.25 sec.
memory limit per test: 65536 KB
input: standard
output: standard



When entering some analog data into a computer, this information must be quantized. Quantization transforms each measured value x to some other value l(x) selected from the predefined set L of levels. Sometimes to reduce the influence of the levels set to the information, the group of levels sets Li is used. The number of levels sets is usually chosen to be the power of 2.

When using the number of levels sets, some additional information should be used to specify which set was used for each quantization. However, providing this information may be too expensive — the better solution would be to choose more levels and use one set. To avoid the specification of the quantization set, the following technique is used. Suppose that n values x1, x2, ..., xn are to be quantized and the group of m=2p levels sets Li, i=0, ..., m-1; each of size s=2q is used to quantize it. After quantization xj is replaced with some number lj = in Lf(j). Instead of sending lj, its ordinal number in Lf(j) is usually sent, let kj be the ordinal number of lj in Lf(j) (levels are numbered starting with 0). Take p least significant bits of kj and say that the number kj & (2p-1) is the number of the levels set that will be used for next quantization, that is f(j+1) = kj & (2p-1).

Since the least significant bits of kj are usually distributed quite randomly, the sets used for quantization change often and weakly depend on values of quantized data, thus this technique provides the good way to perform the quantization.

Usually to perform the quantization the closest to the value level of the levels set is chosen. However, using the technique described, sometimes it pays off to choose not the optimal level, but some other one, the ordinal number of which has other least significant bits, thus choosing another levels set for next measure and providing better approximation of quantized values in the future. Let us call the deviation of quantization the value of sum(j=1.. n, |xj - lj|). Your task is given measures and levels sets to choose quantized value for each measure in such a way, that the deviation of quantization is minimal possible.

The first value is always quantized using set L0.

Input
The first line of the input file contains n (1 ≤ n ≤ 1000). The second line contains n integer numbers xi ranging from 1 to 106. The next line contains m and s (1 ≤ m ≤ 128, m ≤ s ≤ 128). Next m lines contain s integer numbers each — levels of the quantization sets given in increasing order for each set, all levels satisfy 1 ≤ level ≤ 106.

Output
First output the minimal possible deviation of the quantization. Then output n integer numbers in range from 0 to s - 1. For each input value output the number of the level in the corresponding levels set (kj) used for this number to achieve the quantization required.

Sample test(s)

Input
3
8 8 19
2 4
5 10 15 20
3 7 13 17

Output
5
1 1 3
Author:	Andrew Stankevich
Resource:	Petrozavodsk Summer Trainings 2003
Date:	2003-08-23

<|response|>
## 1) Abridged problem statement (concise)

You are given `n` measured integers `x1..xn`. There are `m` quantization level sets `L0..L(m-1)`; each set contains `s` integer levels sorted increasingly.

For each position `j`, you must choose an index `kj` (`0..s-1`) and replace `xj` by the level `Lj[kj]`, where `Lj` is the set used at this position.

The set used changes deterministically:
- The first value uses set `L0`.
- If at position `j` you choose index `kj`, then the next set index is  
  `f(j+1) = kj & (m-1)` (bitwise AND).

Goal: minimize total deviation
\[
\sum_{j=1}^{n} |x_j - L_{f(j)}[k_j]|
\]
Output the minimal deviation and the chosen indices `k1..kn`.

Constraints: `n ≤ 1000`, `m ≤ 128`, `s ≤ 128` (with `m ≤ s`).

---

## 2) Key observations

1. **A choice affects the future**: picking index `k` at step `j` not only incurs immediate error `|xj - Lset[k]|`, but also determines the **next set** via `k & (m-1)`.  
2. This is a typical **dynamic programming over (position, current set)** problem:
   - State: which `x` we are quantizing now, and which set must be used now.
   - Transition: try all possible `k` in the current set; move to next state determined by the low bits of `k`.
3. Constraints are small enough for `O(n * m * s)`:
   - Max: `1000 * 128 * 128 = 16,384,000` transitions.

---

## 3) Full solution approach (DP + reconstruction)

### DP definition
Use 0-based indexing for positions.

Let:
- `dp[j][set]` = minimal total deviation for quantizing values `x[j..n-1]`,
  **given that** value `x[j]` must be quantized using level set `Lset`.

We want `dp[0][0]` because the first value always uses set `0`.

### Base case
If there are no values left:
- `dp[n][set] = 0` for all `set`.

### Transition
At state `(j, set)`, choose any index `k` from `0..s-1`:

- Immediate cost: `abs(x[j] - L[set][k])`
- Next set: `next_set = k & (m - 1)`
- Total cost: `abs(...) + dp[j+1][next_set]`

So:
\[
dp[j][set] = \min_{k=0}^{s-1} \left( |x[j]-L[set][k]| + dp[j+1][(k \& (m-1))] \right)
\]

### Reconstruction (to output `k1..kn`)
Store which `k` achieved the minimum:
- `choice[j][set] = best_k`

Then simulate forward:
- `cur_set = 0`
- for `j = 0..n-1`:
  - `k = choice[j][cur_set]`
  - output `k`
  - `cur_set = k & (m-1)`

### Complexity
- Time: `O(n*m*s)` ≤ ~16.4M simple operations
- Memory: `O(n*m)` for `dp` and `choice`

---

## 4) C++ implementation

```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;
};

const int64_t inf = (int64_t)1e18 + 42;

int n, m, s;
vector<int> x;
vector<vector<int>> L;

void read() {
    cin >> n;
    x.resize(n);
    cin >> x;
    cin >> m >> s;
    L.resize(m, vector<int>(s));
    for(auto& row: L) {
        cin >> row;
    }
}

void solve() {
    // Given n values x[0..n-1], m=2^p level sets each of size s=2^q, where
    // L[i][k] is the k-th level in set i. We must replace each x[j] with some
    // level L[set][k] from the current set, minimizing sum |x[j] - L[set][k]|.
    // The first value uses set 0. The chosen index k's p LSBs
    // determine the next set: next_set = k & (m-1). So picking a suboptimal
    // level now can lead to a better set (and lower total error) later.
    //
    // We solve this with DP backwards. dp[j][set] is the min total deviation
    // for x[j..n-1] when position j uses level set `set`. For each (j, set), we
    // try all s level indices k, pay |x[j] - L[set][k]| and transition to
    // dp[j+1][k & (m-1)]. We track choices for reconstruction. The complexity
    // is O(n * m * s), with n=1000, m=128, s=128 that's ~16M ops, fast enough
    // for the 0.25s time limit.

    vector<vector<int64_t>> dp(n + 1, vector<int64_t>(m, 0));
    vector<vector<int>> choice(n, vector<int>(m, -1));

    for(int j = n - 1; j >= 0; j--) {
        for(int set = 0; set < m; set++) {
            dp[j][set] = inf;
            for(int k = 0; k < s; k++) {
                int next_set = k & (m - 1);
                int64_t cost = abs(x[j] - L[set][k]) + dp[j + 1][next_set];
                if(cost < dp[j][set]) {
                    dp[j][set] = cost;
                    choice[j][set] = k;
                }
            }
        }
    }

    cout << dp[0][0] << '\n';
    int cur_set = 0;
    for(int j = 0; j < n; j++) {
        int k = choice[j][cur_set];
        cout << k << " \n"[j == n - 1];
        cur_set = k & (m - 1);
    }
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 5) Python implementation (detailed comments)

```python
import sys

INF = 10**18

def solve() -> None:
    data = list(map(int, sys.stdin.buffer.read().split()))
    it = iter(data)

    n = next(it)
    x = [next(it) for _ in range(n)]

    m = next(it)
    s = next(it)

    # L[set][k] = k-th level in set 'set'
    L = [[next(it) for _ in range(s)] for _ in range(m)]

    # We'll do backward DP but keep only the next row of dp to save memory:
    # dp_next[set] == dp[j+1][set] for the current j in the backward loop.
    dp_next = [0] * m

    # For reconstruction, we still need choice[j][set].
    choice = [[0] * m for _ in range(n)]

    # Iterate positions from n-1 down to 0
    for j in range(n - 1, -1, -1):
        dp_cur = [INF] * m
        xj = x[j]

        for set_idx in range(m):
            best_cost = INF
            best_k = 0
            levels = L[set_idx]

            # Try all possible indices k
            for k in range(s):
                next_set = k & (m - 1)
                cost = abs(xj - levels[k]) + dp_next[next_set]
                if cost < best_cost:
                    best_cost = cost
                    best_k = k

            dp_cur[set_idx] = best_cost
            choice[j][set_idx] = best_k

        dp_next = dp_cur  # move dp row upward

    # dp_next[0] is dp[0][0] after finishing the loop
    min_deviation = dp_next[0]

    # Reconstruct the sequence of chosen indices k1..kn
    cur_set = 0
    ks = []
    for j in range(n):
        k = choice[j][cur_set]
        ks.append(k)
        cur_set = k & (m - 1)

    sys.stdout.write(str(min_deviation) + "\n")
    sys.stdout.write(" ".join(map(str, ks)) + "\n")

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

