#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

const ll MOD = 1e9 + 7;

mat mul(mat& A, mat& B) {
	mat C(A.size(), vec(B[0].size(), 0));
	for (int i = 0; i < A.size(); i++) {
		for (int k = 0; k < B.size(); k++) {
			for (int j = 0; j < B[0].size(); j++) {
				(C[i][j] += A[i][k] * B[k][j]) %= MOD; 
			}
		}
	}
	return C;
}

mat pow(mat A, ll n) {
	mat B(A.size(), vec(A.size()));
	for (int i = 0; i < A.size(); i++) {
		B[i][i] = 1;
	}
	while (n > 0) {
		if (n & 1) B = mul(B, A);
		A = mul(A, A);
		n >>= 1;
	}
	return B;
}


int N;
ll K, a[10000];

void solve1() {
	ll f = 0, s = 0;
	queue<ll> q;
	for (int i = 0; i < N; i++) {
		f += a[i];
		s += a[i];
		q.push(a[i]);
	}
	s += f;
	q.push(f);
	for (int i = N + 1; i < K; i++) {
		f = (f * 2 - q.front() + MOD) % MOD;
		q.push(f);
		(s += f) %= MOD;
		q.pop();
	}
	cout << f << " " << s << endl;
}

void solve2() {
	mat A(N + 1, vec(N + 1, 0));
	for (int i = 0; i <= N; i++) A[0][i] = 1;
	for (int i = 1; i <= N; i++) A[1][i] = 1;
	for (int i = 2; i <= N; i++) A[i][i - 1] = 1;
	A = pow(A, K - N);
	mat B(N + 1, vec(1, 0));
	for (int i = 0; i < N; i++) {
		B[N - i][0] = a[i];
		B[0][0] += a[i];
	}
	A = mul(A, B);
	cout << A[1][0] << " " << A[0][0] << endl;
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> N >> K;
	for (int i = 0; i < N; i++) cin >> a[i];
	if (K <= 1000000) solve1();
	else solve2();
	return 0;
}