Problem:
Score : 300 points

Problem

Takahashi, who is a novice in competitive programming, wants to learn M algorithms.
Initially, his understanding level of each of the M algorithms is 0.

Takahashi is visiting a bookstore, where he finds N books on algorithms.
The i-th book (1\leq i\leq N) is sold for C_i yen (the currency of Japan). If he buys and reads it, his understanding level of the j-th algorithm will increase by A_{i,j} for each j (1\leq j\leq M).
There is no other way to increase the understanding levels of the algorithms.

Takahashi's objective is to make his understanding levels of all the M algorithms X or higher. Determine whether this objective is achievable. If it is achievable, find the minimum amount of money needed to achieve it.

Constraints

All values in input are integers.

1\leq N, M\leq 12

1\leq X\leq 10^5

1\leq C_i \leq 10^5

0\leq A_{i, j} \leq 10^5

Input

Input is given from Standard Input in the following format:

N M X
C_1 A_{1,1} A_{1,2} \cdots A_{1,M}
C_2 A_{2,1} A_{2,2} \cdots A_{2,M}
\vdots
C_N A_{N,1} A_{N,2} \cdots A_{N,M}

Output

If the objective is not achievable, print -1; otherwise, print the minimum amount of money needed to achieve it.

Sample Input 1

3 3 10
60 2 2 4
70 8 7 9
50 2 3 9

Sample Output 1

120

Buying the second and third books makes his understanding levels of all the algorithms 10 or higher, at the minimum cost possible.

Sample Input 2

3 3 10
100 3 1 4
100 1 5 9
100 2 6 5

Sample Output 2

-1

Buying all the books is still not enough to make his understanding levels of all the algorithms 10 or higher.

Sample Input 3

8 5 22
100 3 7 5 3 1
164 4 5 2 7 8
334 7 2 7 2 9
234 4 7 2 8 2
541 5 4 3 3 6
235 4 8 6 9 7
394 3 6 1 6 2
872 8 4 3 7 2

Sample Output 3

1067

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

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

typedef long long ll;
#define pa pair<int, int>
#define pal pair<long long, long long>
#define pali pair<long long, int>
#define pad pair<double, double>
#define pb push_back
#define mp make_pair
#define COUT(v)                              \
    for (int64_t i = 0; i < (v).size(); ++i) \
    {                                        \
        cout << v.at(i) << endl;             \
    }
#define REP(i, n) for (int64_t i = 0; i < n; ++i)
#define FOR(i, r, n) for (int64_t i = (r); i < n; ++i)
#define VIN(v)                               \
    for (int64_t i = 0; i < (v).size(); ++i) \
    {                                        \
        cin >> (v).at(i);                    \
    }

typedef vector<bool> bvec;
typedef vector<int> ivec;
typedef vector<long long> lvec;
typedef vector<double> dvec;
typedef vector<pa> pavec;
typedef vector<pali> palivec;
typedef vector<pal> palvec;

typedef vector<vector<bool>> bmat;
typedef vector<vector<int>> imat;
typedef vector<vector<long long>> lmat;

typedef vector<string> svec;
typedef vector<vector<string>> smat;
const ll infll = (1LL << 60) - 1;
const int inf = (1 << 30) - 1;
const ll MOD = 1000000007;

ll gcd(ll x, ll y)
{
    ll r = x % y;
    if (r == 0)
        return y;
    else
        return gcd(y, r);
}

ll lcm(ll x, ll y)
{
    return x * y / gcd(x, y);
}

lvec mfactor(ll n)
{
    bvec ip(n, true);
    lvec mf(n, -1);
    ip[0] = false;
    ip[1] = false;
    mf[0] = 0;
    mf[1] = 1;
    REP(i, n)
    {
        if (ip[i])
        {
            mf[i] = i;
            for (ll j = i * i; j < n; j += i)
            {
                ip[j] = false;
                if (mf[j] == -1)
                    mf[j] = i;
            }
        }
    }
    return mf;
}

palivec get_prime(ll n, const lvec &mf)
{
    palivec plist;
    while (n != 1)
    {
        int cnt = 0;
        ll m = mf[n];
        while (mf[n] == m)
        {
            cnt++;
            n /= m;
        }
        plist.pb(pali(m, cnt));
    }
    return plist;
}

void COMinit(int m, lvec &fac, lvec &finv)
{
    lvec inv(m);
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < m; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

ll pow_mod(ll a, ll n)
{
    ll x = 1;
    while (n > 0)
    {
        if (n & 1)
        {
            x = x * a % MOD;
        }
        a = a * a % MOD;
        n >>= 1;
    }
    return x;
}

ll pow_mod2(ll a, ll n)
{
    ll x = 1;
    while (n > 0)
    {
        if (n & 1)
        {
            x = x * a;
        }
        a = a * a;
        n >>= 1;
    }
    return x;
}

ll COM(int n, int k, const lvec &fac, const lvec &finv)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return (fac[n] * (finv[k] * finv[n - k] % MOD)) % MOD;
}

ll modinv(ll a, ll m)
{
    ll b = m, u = 1, v = 0;
    while (b)
    {
        ll t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0)
        u += m;
    return u;
}

// union by size + path having
class UnionFind
{
public:
    vector<ll> par;
    vector<ll> siz;

    // Constructor
    UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL)
    {
        for (ll i = 0; i < sz_; ++i)
            par[i] = i;
    }
    void init(ll sz_)
    {
        par.resize(sz_);
        siz.assign(sz_, 1LL);
        for (ll i = 0; i < sz_; ++i)
            par[i] = i;
    }

    // Member Function
    // Find
    ll root(ll x)
    {
        while (par[x] != x)
        {
            x = par[x] = par[par[x]];
        }
        return x;
    }

    // Union(Unite, Merge)
    bool merge(ll x, ll y)
    {
        x = root(x);
        y = root(y);
        if (x == y)
            return false;
        if (siz[x] < siz[y])
            swap(x, y);
        siz[x] += siz[y];
        par[y] = x;
        return true;
    }

    bool issame(ll x, ll y)
    {
        return root(x) == root(y);
    }

    ll size(ll x)
    {
        return siz[root(x)];
    }
};

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, m, x;
    cin >> n >> m >> x;
    ivec c(n);
    imat a(n, ivec(m));

    for (int i = 0; i < n; i++)
    {
        cin >> c[i];
        for (int j = 0; j < m; j++)
        {
            cin >> a[i][j];
        }
    }
    int inf = 1e7;
    int ans = inf;
    int v = 0;
    function<bool(int, int)> check = [&](int v, int d) {
        ivec xv(m, 0);
        for (int i = 0; i <= d; i++)
        {
            if ((v >> i) & 1)
            {
                for (int j = 0; j < m; j++)
                {
                    xv[j] += a[i][j];
                }
            }
        }

        for (auto vv : xv)
        {
            if (vv < x)
            {
                return false;
            }
        }
        //return all_of(xv.begin(), xv.end(), [&](int xx) { return xx >= x; });
        return true;
        /*
        for (int i = 0; i < m; i++)
        {
            if (xv[i] < x)
            {
                return false;
            }
        }
        return true;
        */
    };

    function<void(int, int)> dfs = [&](int d, int s) {
        if (d == n)
        {
            return;
        }
        v &= ~(1 << d);
        dfs(d + 1, s);
        v |= 1 << d;
        if (check(v, d))
        {
            ans = min(ans, s + c[d]);
        }
        else
        {
            dfs(d + 1, s + c[d]);
        }
    };
    dfs(0, 0);
    if (ans == inf)
    {
        cout << -1 << endl;
    }
    else
    {
        cout << ans << endl;
    }
}