#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using P = pair<ll, ll>;
using Graph = vector<vector<ll>>;
#define rep(i, n) for(ll i=0;i<(ll)(n);i++)
#define rep2(i, m, n) for(ll i=m;i<(ll)(n);i++)
#define rrep(i, n, m) for(ll i=n;i>=(ll)(m);i--)
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int ddx[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll MOD = 1000000007;
const ll INF = 1000000000000000000L;
#ifdef __DEBUG

/**
 * For DEBUG
 * https://github.com/ta7uw/cpp-pyprint
 */
#include "cpp-pyprint/pyprint.h"

#endif

class mint {
public:
    long long x;

    constexpr mint(long long x = 0) : x((x % MOD + MOD) % MOD) {}

    constexpr mint operator-() const {
        return mint(-x);
    }

    constexpr mint &operator+=(const mint &rhs) {
        if ((x += rhs.x) >= MOD) x -= MOD;
        return *this;
    }

    constexpr mint &operator-=(const mint &rhs) {
        if ((x += MOD - rhs.x) >= MOD) x -= MOD;
        return *this;
    }

    constexpr mint &operator*=(const mint &rhs) {
        (x *= rhs.x) %= MOD;
        return *this;
    }

    constexpr mint operator+(const mint &rhs) const {
        mint res(*this);
        return res += rhs;
    }

    constexpr mint operator-(const mint &rhs) const {
        mint res(*this);
        return res -= rhs;
    }

    constexpr mint operator*(const mint &rhs) const {
        mint res(*this);
        return res *= rhs;
    }

    [[nodiscard]] constexpr mint pow(long long t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    [[nodiscard]] constexpr mint inv() const {
        return pow(MOD - 2);
    }

    mint &operator/=(mint rhs) {
        return *this *= rhs.inv();
    }

    mint operator/(const mint rhs) const {
        return mint(*this) /= rhs;
    }

    constexpr bool operator==(const mint &rhs) const noexcept {
        return this->x == rhs.x;
    }

    constexpr bool operator!=(const mint &rhs) const noexcept {
        return this->x != rhs.x;
    }

    bool operator<(const mint &rhs) const noexcept {
        return this->x < rhs.x;
    }

    bool operator>(const mint &rhs) const noexcept {
        return this->x > rhs.x;
    }

    bool operator<=(const mint &rhs) const noexcept {
        return this->x <= rhs.x;
    }

    bool operator>=(const mint &rhs) const noexcept {
        return this->x >= rhs.x;
    }

    friend istream &operator>>(istream &is, mint &a) {
        long long t;
        is >> t;
        a = mint(t);
        return is;
    }

    friend ostream &operator<<(ostream &os, const mint &a) {
        return os << a.x;
    }
};


mint maxProductSubarrayOfSizeK(vector<ll> &A, ll n, ll k) {
    // sorting given input array
    sort(A.begin(), A.end());

    // variable to store final product of all element
    // of sub-sequence of size k
    mint product = 1;

    // CASE I
    // If max element is 0 and
    // k is odd then max product will be 0
    if (A[n - 1] == 0 && (k & 1))
        return 0;

    // CASE II
    // If all elements are negative and
    // k is odd then max product will be
    // product of rightmost-subarray of size k
    if (A[n - 1] < 0 && (k & 1)) {
        for (int i = n - 1; i >= n - k; i--)
            product *= A[i];
        return product;
    }

    // else
    // i is current left pointer index
    int i = 0;

    // j is current right pointer index
    int j = n - 1;

    // CASE III
    // if k is odd and rightmost element in
    // sorted array is positive then it
    // must come in subsequence
    // Multiplying A[j] with product and
    // correspondingly changing j
    if (k & 1) {
        product *= A[j];
        j--;
        k--;
    }

    // CASE IV
    // Now k is even
    // Now we deal with pairs
    // Each time a pair is multiplied to product
    // ie.. two elements are added to subsequence each time
    // Effectively k becomes half
    // Hence, k >>= 1 means k /= 2
    k >>= 1;

    // Now finding k corresponding pairs
    // to get maximum possible value of product
    for (int itr = 0; itr < k; itr++) {

        // product from left pointers
        ll left_product = A[i] * A[i + 1];

        // product from right pointers
        ll right_product = A[j] * A[j - 1];

        // Taking the max product from two choices
        // Correspondingly changing the pointer's position
        if (left_product > right_product) {
            product *= left_product;
            i += 2;
        } else {
            product *= right_product;
            j -= 2;
        }
    }

    // Finally return product
    return product;
}

void solve() {
    ll N, K;
    cin >> N >> K;
    vector<ll> A(N);
    rep(i, N) cin >> A[i];
    cout << maxProductSubarrayOfSizeK(A, N, K) << '\n';
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    solve();
    return 0;
}


Measured memory_bytes: 4748