Problem: <image>

William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.

For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.

Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.

Input

The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.

Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.

Output

Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.

If there are multiple answers, print any.

Examples

Input


3 4 3
1000
0110
1001


Output


1000


Input


5 5 4
11001
10101
10010
01110
11011


Output


10001

Note

In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.

In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

// template {{{
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define lb lower_bound
#define ub upper_bound
#define f first
#define s second
#define resz resize

#define sz(x) int((x).size())
#define all(x) (x).begin(), (x).end()

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i, a, b) for (int i = (b)-1; i >= (a); i--)
#define F0Rd(i, a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto& a : x)

#define sort_by(x, y) sort(all(x), [&](const auto& a, const auto& b) { return y; })

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;

using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;

using vpii = vector<pii>;
using vpll = vector<pll>;
using vpdd = vector<pdd>;

template<typename T> void ckmin(T& a, const T& b) { a = min(a, b); }
template<typename T> void ckmax(T& a, const T& b) { a = max(a, b); }

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

namespace __input {
    template<class T1, class T2> void re(pair<T1,T2>& p);
    template<class T> void re(vector<T>& a);
    template<class T, size_t SZ> void re(array<T,SZ>& a);

    template<class T> void re(T& x) { cin >> x; }
    void re(double& x) { string t; re(t); x = stod(t); }
    template<class Arg, class... Args> void re(Arg& first, Args&... rest) {
        re(first); re(rest...);
    }

    template<class T1, class T2> void re(pair<T1,T2>& p) { re(p.f,p.s); }
    template<class T> void re(vector<T>& a) { F0R(i,sz(a)) re(a[i]); }
    template<class T, size_t SZ> void re(array<T,SZ>& a) { F0R(i,SZ) re(a[i]); }
}
using namespace __input;

namespace __output {
    template<class T1, class T2> void pr(const pair<T1,T2>& x);
    template<class T, size_t SZ> void pr(const array<T,SZ>& x);
    template<class T> void pr(const vector<T>& x);
    template<class T> void pr(const set<T>& x);
    template<class T1, class T2> void pr(const map<T1,T2>& x);

    template<class T> void pr(const T& x) { cout << x; }
    template<class Arg, class... Args> void pr(const Arg& first, const Args&... rest) {
        pr(first); pr(rest...);
    }

    template<class T1, class T2> void pr(const pair<T1,T2>& x) {
        pr("{",x.f,", ",x.s,"}");
    }
    template<class T, bool pretty = true> void prContain(const T& x) {
        if (pretty) pr("{");
        bool fst = 1; for (const auto& a: x) pr(!fst?pretty?", ":" ":"",a), fst = 0;
        if (pretty) pr("}");
    }
    template<class T> void pc(const T& x) { prContain<T, false>(x); pr("\n"); }
    template<class T, size_t SZ> void pr(const array<T,SZ>& x) { prContain(x); }
    template<class T> void pr(const vector<T>& x) { prContain(x); }
    template<class T> void pr(const set<T>& x) { prContain(x); }
    template<class T1, class T2> void pr(const map<T1,T2>& x) { prContain(x); }

    void ps() { pr("\n"); }
    template<class Arg> void ps(const Arg& first) {
        pr(first); ps();
    }
    template<class Arg, class... Args> void ps(const Arg& first, const Args&... rest) {
        pr(first," "); ps(rest...);
    }
}
using namespace __output;

#define TRACE(x) x
#define __pn(x) pr(#x, " = ")
#define pd(...) __pn((__VA_ARGS__)), ps(__VA_ARGS__), cout << flush

namespace __algorithm {
    template<typename T> void dedup(vector<T>& v) {
        sort(all(v)); v.erase(unique(all(v)), v.end());
    }
    template<typename T> typename vector<T>::const_iterator find(const vector<T>& v, const T& x) {
        auto it = lower_bound(all(v), x); return it != v.end() && *it == x ? it : v.end();
    }
    template<typename T> size_t index(const vector<T>& v, const T& x) {
        auto it = find(v, x); assert(it != v.end() && *it == x); return it - v.begin();
    }

    template<typename T1, typename T2> typename vector<pair<T1, T2>>::iterator lower_bound(
            const vector<pair<T1, T2>>& v, const T1& x) {
        return lower_bound(all(v), x, [](pair<T1, T2> a, pair<T1, T2> b) { return a.f < b.f; });
    }
    template<typename T1, typename T2> typename vector<pair<T1, T2>>::iterator upper_bound(
            const vector<pair<T1, T2>>& v, const T1& x) {
        return upper_bound(all(v), x, [](pair<T1, T2> a, pair<T1, T2> b) { return a.f < b.f; });
    }
}
using namespace __algorithm;

struct __monostate {
    friend istream& operator>>(istream& is, const __attribute__((unused))__monostate& ms) { return is; }
    friend ostream& operator<<(ostream& os, const __attribute__((unused))__monostate& ms) { return os; }
} ms;

namespace __io {
    FILE* setIn(string s) { return freopen(s.c_str(),"r",stdin); }
    FILE* setOut(string s) { return freopen(s.c_str(),"w",stdout); }
    void setIO(string s = "") {
        ios_base::sync_with_stdio(0); cin.tie(0);
        cout << setprecision(15);
        if (sz(s)) { setIn(s+".in"), setOut(s+".out"); }
    }
}
using namespace __io;
// }}}

int main() {
    setIO();

    int N, M, P;
    re(N, M, P);

    vector<int64_t> liked(N);

    F0R (i, N) {
        F0R (j, M) {
            char c; re(c);
            if (c == '1') {
                liked[i] |= int64_t(1) << j;
            }
        }
    }

    const int cutoff = (N + 1) / 2;

    pair<int, int64_t> ans = { 0, 0 };

    F0R (iter, 50) {
        int p = rng() % N;

        vi has;
        F0R (j, M) {
            if ((liked[p] >> j)&1)
                has.pb(j);
        }

        const int H = sz(has);
        assert(H <= P);

        vi super(1 << H);
        F0R (i, N) {
            int mask = 0;
            F0R (h, H) { // 50 * 2e5 * 15
                if ((liked[i] >> has[h])&1)
                    mask |= 1 << h;
            }
            super[mask] += 1;
        }

        F0R (h, H) {
            F0R (m, 1 << H) { // 50 * 15 * 2^15
                if (!(m & (1 << h))) {
                    super[m] += super[m | (1 << h)];
                }
            }
        }

        F0R (m, 1 << H) {
            int pop = __builtin_popcount(m);
            if (pop > ans.f && super[m] >= cutoff) {
                ans = { pop, 0 };
                F0R (h, H) { // 50 * 2^15 * 15
                    if (m & (1 << h)) {
                        ans.s |= int64_t(1) << has[h];
                    }
                }
            }
        }
    }

    F0R (j, M) {
        pr((ans.s >> j)&1);
    }
    ps();

    return 0;
}