// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define F first

#define S second

#define pii pair<int, int>

#define eb emplace_back

#define all(v) v.begin(), v.end()

#define rep(i, n) for (int i = 0; i < (n); ++i)

#define rep3(i, l, n) for (int i = l; i < (n); ++i)

#define sz(v) (int)v.size()

#define inf (int)(1e9+7)

#define INF (ll)(1e18)

#define abs(x) (x >= 0 ? x : -(x))

#define ceil(a, b) a / b + !!(a % b)

template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }

template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }





template<typename T>

class SegTree {

    int n;

    T def;

    vector<T> data;

    function<T(T, T)> operation;

    function<T(T, T)> update;

    

    T _query(int a, int b, int k, int l, int r) {

        if (r <= a || l >= b) return def;

        if (a <= l && r <= b) return data[k];

        T c1 = _query(a, b, 2 * k + 1, l, (l + r) / 2);

        T c2 = _query(a, b, 2 * k + 2, (l + r) / 2, r);

        return operation(c1, c2);

    }

    

public:

    SegTree<T>(int _n, T _def, function<T(T, T)> _operation, function<T(T, T)> _update)

            : def(_def), operation(_operation), update(_update) {

                n = 1;

                while (n < _n) n *= 2;

                data = vector<T>(2 * n - 1, def);

            }

    

    void change(int i, T x) {

        i += n - 1;

        data[i] = update(data[i], x);

        while (i > 0) {

            i = (i - 1) / 2;

            data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);

        }

    }

    

    T query(int a, int b) {

        return _query(a, b, 0, 0, n);

    }

    

    T operator[](int i) { return data[i + n - 1]; }

};



int main() {

    ios::sync_with_stdio(false);

    cin.tie(0);

    

    

    int n, m;

    cin >> n >> m;

    string s;

    cin >> s;

    

    SegTree<int> sgt(n + 1, inf,

                     [](int a, int b) { return min(a, b); },

                     [](int a, int b) { return b; });

    sgt.change(n, 0);

    

    for (int i = n - 1; i >= 0; --i) {

        if (s[i] == '1') continue;

        int mi = sgt.query(i + 1, min(i + m + 1, n + 1));

        if (mi != inf) sgt.change(i, mi + 1);

    }

    // rep(i, n + 1) cout << sgt[i] << " "; cout << endl;

    

    if (sgt[0] == inf) { cout << -1 << endl; return 0; }

    

    vector<int> ans;

    ans.eb(0);

    int table[200005] = {0};

    table[sgt[0]] = 1;

    

    rep3(i, 1, n + 1) {

        if (sgt[i] != sgt[i - 1] && sgt[i] != inf && table[sgt[i]] == 0) {

            ans.eb(i);

            table[sgt[i]] = 1;

        }

    }

    

    // rep(i, sz(ans)) cout << ans[i] << " "; cout << endl;

    rep3(i, 1, sz(ans)) cout << ans[i] - ans[i - 1] << " "; cout << endl;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define F first

#define S second

#define pii pair<int, int>

#define eb emplace_back

#define all(v) v.begin(), v.end()

#define rep(i, n) for (int i = 0; i < (n); ++i)

#define rep3(i, l, n) for (int i = l; i < (n); ++i)

#define sz(v) (int)v.size()

const int inf = 1e9 + 7;

const ll INF = 1e18;

int mod = 1000000007;

#define abs(x) (x >= 0 ? x : -(x))

#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())

#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())

template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }

template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }

template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }

template<typename T> T pow(T a, int b) { return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1; }

ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }

template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { for (auto &vi: vec) os << vi << " "; return os; }

template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }

template<typename T >inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }

template<class T> inline void add(T &a, int b) { a += b; if (a >= mod) a -= mod; }





void solve();



int main() {

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    cout << fixed << setprecision(10);



    int T;

    // cin >> T;

    T = 1;



    while (T--) {

        solve();

    }

}



void solve() {

  int n, m;

  cin >> n >> m;

  string s;

  cin >> s;



  int now = n;

  vector<int> ans;



  while (1) {

    for (int j = m; j >= 1; --j) {

      if (now - j >= 0 && s[now - j] == '0') {

        // cout << now << " " << j << endl;

        now -= j;

        ans.eb(j);

        if (now == 0) {

          reverse(all(ans));

          cout << ans << endl;

          return;

        }

        break;

      }

      if (j == 1) {

        cout << -1 << endl;

        return;

      }

    }

  }

}


// Fastest known solution for this problem: