description: The Winter holiday will be here soon. Mr. Chanek wants to decorate his house's wall with ornaments. The wall can be represented as a binary string a of length n. His favorite nephew has another binary string b of length m (m ≤ n).
  
  Mr. Chanek's nephew loves the non-negative integer k. His nephew wants exactly k occurrences of b as substrings in a. 
  
  However, Mr. Chanek does not know the value of k. So, for each k (0 ≤ k ≤ n - m + 1), find the minimum number of elements in a that have to be changed such that there are exactly k occurrences of b in a.
  
  A string s occurs exactly k times in t if there are exactly k different pairs (p,q) such that we can obtain s by deleting p characters from the beginning and q characters from the end of t.
  
  Input
  
  The first line contains two integers n and m (1 ≤ m ≤ n ≤ 500) — size of the binary string a and b respectively.
  
  The second line contains a binary string a of length n.
  
  The third line contains a binary string b of length m.
  
  Output
  
  Output n - m + 2 integers — the (k+1)-th integer denotes the minimal number of elements in a that have to be changed so there are exactly k occurrences of b as a substring in a.
  
  Example
  
  Input
  
  
  9 3
  100101011
  101
  
  
  Output
  
  
  1 1 0 1 6 -1 -1 -1
  
  Note
  
  For k = 0, to make the string a have no occurrence of 101, you can do one character change as follows.
  
  100101011 → 100100011
  
  For k = 1, you can also change a single character.
  
  100101011 → 100001011
  
  For k = 2, no changes are needed.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int MOD = (int)1e9 + 7;
  const int FFTMOD = 119 << 23 | 1;
  const int INF = (int)1e9 + 23111992;
  const long long LINF = (long long)1e18 + 23111992;
  const long double PI = acos((long double)-1);
  const long double EPS = 1e-9;
  inline long long gcd(long long a, long long b) {
    long long r;
    while (b) {
      r = a % b;
      a = b;
      b = r;
    }
    return a;
  }
  inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  inline long long fpow(long long n, long long k, int p = MOD) {
    long long r = 1;
    for (; k; k >>= 1) {
      if (k & 1) r = r * n % p;
      n = n * n % p;
    }
    return r;
  }
  template <class T>
  inline int chkmin(T& a, const T& val) {
    return val < a ? a = val, 1 : 0;
  }
  template <class T>
  inline int chkmax(T& a, const T& val) {
    return a < val ? a = val, 1 : 0;
  }
  inline unsigned long long isqrt(unsigned long long k) {
    unsigned long long r = sqrt(k) + 1;
    while (r * r > k) r--;
    return r;
  }
  inline long long icbrt(long long k) {
    long long r = cbrt(k) + 1;
    while (r * r * r > k) r--;
    return r;
  }
  inline void addmod(int& a, int val, int p = MOD) {
    if ((a = (a + val)) >= p) a -= p;
  }
  inline void submod(int& a, int val, int p = MOD) {
    if ((a = (a - val)) < 0) a += p;
  }
  inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
  inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
  inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
  inline int sign(long double x, long double y) { return sign(x - y); }
  mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
  inline int mrand() { return abs((int)mt()); }
  inline int mrand(int k) { return abs((int)mt()) % k; }
  struct KMP {
    string s;
    int n;
    vector<int> link;
    void build(string str) {
      s = str;
      n = int((s).size());
      link.resize(n);
      int cur = link[0] = -1;
      for (int i = 1; i < n; i++) {
        while (cur >= 0 && s[i] != s[cur + 1]) cur = link[cur];
        if (s[i] == s[cur + 1]) cur++;
        link[i] = cur;
      }
    }
    vector<int> query(char* t) {
      vector<int> res;
      int k = strlen(t);
      int cur = -1;
      for (int i = 0; i < k; i++) {
        while (cur >= 0 && t[i] != s[cur + 1]) cur = link[cur];
        if (t[i] == s[cur + 1]) cur++;
        if (cur == n - 1) {
          res.push_back(i - n + 1);
          cur = link[cur];
        }
      }
      return res;
    }
  } kmp;
  void chemthan() {
    int n, m;
    cin >> n >> m;
    string a, b;
    cin >> a >> b;
    kmp.build(b);
    vector<vector<int>> mem(m + 5, vector<int>(2, -2));
    auto next_state = [&](int v, int p) {
      int& res = mem[v + 1][p];
      if (res != -2) return res;
      res = v;
      while (0 <= res && b[res + 1] != '0' + p) res = kmp.link[res];
      if (b[res + 1] == '0' + p) res++;
      return res;
    };
    vector<vector<vector<int>>> dp(
        2, vector<vector<int>>(m + 5, vector<int>(n - m + 5, INF)));
    int cur = 0, prv = 1;
    for (int u = (n + 1) - 1; u >= (0); --u) {
      for (int v = (0); v < (m + 1); ++v)
        for (int c = (0); c < (n - m + 2); ++c) dp[cur][v][c] = INF;
      for (int v = (-1); v < (m - 1); ++v)
        for (int c = (0); c < (n - m + 2); ++c) {
          int& res = dp[cur][v + 1][c];
          if (u == n) {
            if (!c) {
              res = 0;
            }
            continue;
          }
          for (int i = (0); i < (2); ++i) {
            int cost = a[u] != '0' + i;
            int nv = next_state(v, i);
            int nc = c;
            if (nv == m - 1) {
              nc--;
              nv = kmp.link[nv];
            }
            if (0 <= nc) {
              chkmin(res, dp[prv][nv + 1][nc] + cost);
            }
          }
        }
      swap(cur, prv);
    }
    for (int i = (0); i < (n - m + 2); ++i) {
      int res = dp[prv][0][i];
      if (n < res) res = -1;
      cout << res << " \n"[i == n - m + 1];
    }
  }
  int32_t main(int32_t argc, char* argv[]) {
    ios_base::sync_with_stdio(0), cin.tie(0);
    if (argc > 1) {
      assert(freopen(argv[1], "r", stdin));
    }
    if (argc > 2) {
      assert(freopen(argv[2], "wb", stdout));
    }
    chemthan();
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
  }
  