#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;

int a[N];

struct op {
    int v, ud, val, tlen, len;
    op *prev;

    op(int V = 0, int UD = 0, int VAL = 0, int TLEN = 0, int LEN = 0, op *PREV = NULL) {
        v = V, ud = UD, val = VAL, tlen = TLEN, len = LEN, prev = PREV;
    }

    int operator < (op y) {
        if (v != y.v) return v < y.v;
		if (len != y.len) return len < y.len;
        if (tlen != y.tlen) return tlen < y.tlen;
        if (ud != y.ud) return ud < y.ud;
        if (val != y.val) return val < y.val;
        return 0;
    }

    int operator > (op y) {
        return y < *this;
    }

    int operator == (op y) {
        return !(*this < y) && !(*this > y);
    }
};

vector<op> f[N];
int res[N];

signed main() {
    int n, m; scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    f[1].push_back(op(1, 1, -1, 1, 1));
    f[1].push_back(op(0, 1, 0, 1, 1));
    f[1].push_back(op(0, -1, 0, 1, 1));
    f[1].push_back(op(1, -1, 1, 1, 1));
    for (int i = 2; i <= n; ++i) {
        for (auto &x : f[i - 1]) {
            if (x.ud != -1) f[i].push_back(op(x.v + 1, -1, -1, 1, x.tlen + 1, &x));
            if (x.ud != 1) f[i].push_back(op(x.v + 1, 1, 1, 1, x.tlen + 1, &x));
            int V = (x.val == -1 ? 0 : (!x.val ? a[i - 1] : 1e5)),
                ud = (a[i] == V ? x.ud : (a[i] < V ? -1 : 1)),
                tlen = (a[i] == V ? x.tlen + 1 : 1),
                len = (a[i] == V || ((a[i] > V) ^ (x.ud == -1)) ? x.len + 1 : x.tlen + 1);
            f[i].push_back(op(x.v, ud, 0, tlen, len, &x));
        }
        map< pair<int, int>, op> M;
        for (auto x : f[i]) if (x.len < m && (i == n || x.tlen + 1 < m)) {
            if (!M.count(make_pair(x.ud, x.val)) || M[make_pair(x.ud, x.val)] > x) {
                M[make_pair(x.ud, x.val)] = x;
            }
        }
        f[i].clear();
        for (auto [x, y] : M) f[i].push_back(y);
    }
    op now = *min_element(f[n].begin(), f[n].end());
    printf("%d\n", now.v);
    for (int i = n; i >= 1; --i) {
        res[i] = (now.val == -1 ? 0 : (!now.val ? a[i] : 1e5));
        if (i != 1) now = *(now.prev);
    }
    for (int i = 1; i <= n; ++i) printf("%d%c", res[i], " \n"[i == n]);
    return 0;
}