// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 1.7869507583194433x):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define F first

#define S second

#define pii pair<int, int>

#define pli pair<ll, int>

#define pil pair<int, ll>

#define pll pair<ll, ll>

#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 endl '\n'

const int inf = 1000000007;

const ll INF = 1e18;

// int mod = 998244353;

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; }

ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b); }

ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

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

template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { rep(i, sz(vec)) { if (i) os << " "; os << vec[i]; } 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, class T2> inline void add(T &a, T2 b) { a += b % mod; if (a >= mod) a -= mod; }

template<class T> void operator += (vector<T>& v, vector<T> v2) { rep(i, sz(v2)) v.eb(v2[i]); }



void solve();



int main() {

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    cout << fixed << setprecision(12); // debuged

    int T;

    T = 1;

    while (T--) solve();

}



template<typename T>

class SegTree {

  int n;

  T def;

  vector<T> data;

  function<T(T, T)> operation; // query で

  function<T(T, T)> update; // change で

 

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

    if (r <= a || b <= l) 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);

    }

 

  T query(int a, int b) {

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

  }

 

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

  void out() { rep(i, n) cout << data[i + n - 1] << " "; cout << endl; }

};

// SegTree<int> sgt(n, inf, [](int a, int b) { return a + b; /* query */ }, [](int a, int b) { return b; /* change */ });



// https://atcoder.jp/contests/abc146/submissions/15297851

void solve() {

    int n, m;

    cin >> n >> m;

    string s;

    cin >> s;

    

    SegTree<int> dp(n + 1, inf, [](int a, int b) { return min(a, b); /* query */ }, [](int a, int b) { return b; /* change */ });

    dp.change(n, 0);

    for (int i = n - 1; i >= 0; --i) { // debuged, i=n-2 to n-1

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

        dp.change(i, dp.query(i + 1, i + m + 1) + 1);

    }

    // dp.out();

    if (dp.query(0, 1) == inf) {

        cout << -1 << endl;

        return;

    }

    vector<int> ans;

    int now = n;

    while (now != 0) {

        for (int i = now - m; i <= now - 1; ++i) {

            if (dp.query(i, i + 1) == dp.query(now, now + 1) + 1) {

                ans.eb(now - i);

                now = i;

                break;

            }

        }

    }

    reverse(all(ans));

    cout << ans << endl;

}