# -*- coding: utf-8 -*-
# 想定解(2)

def matmult(A,B): # 正方行列A*B
	n=len(A)
	C=[[0 for i in range(n)] for j in range(n)]
	for x in range(n):
		for z in range(n):
			for y in range(n):
				C[x][y] += A[x][z]*B[z][y]
				C[x][y] %= mo;
	return list(C)

def matpow(A,p): # 正方行列A^p
	n=len(A)
	A=list(A)
	R=[[0 for i in range(n)] for j in range(n)]
	for i in range(n):
		R[i][i]=1
	while p:
		if p%2:
			R = matmult(A,R)
		A=matmult(A,A)
		p >>= 1
	return R

def GetSum(A,x):
	# S[x]を求める
	
	# 行列累乗
	Ap = matpow(A,x-N)
	ret = 0
	for i in range(N+1):
		ret += S[N-i] * Ap[0][i]
	return ret % mo

N,K = map(int, raw_input().strip().split())
F = map(int, raw_input().strip().split())
F.insert(0,0)

mo = 1000000007

S = [0]
for i in range(1,N+1):
	S.append((S[i-1]+F[i]) % mo)

if N > 50:
	# こちらは想定解(1)と同じです
	for i in range(N+1,K+1):
		F.append((S[i-1]-S[i-N-1]) % mo)
		S.append((S[i-1]+F[i]) % mo)
	
	print "%d %d" % (F[K], S[K])
	
else:
	# Sに関する行列累乗を使うケース
	A=[[0 for i in range(N+1)] for j in range(N+1)]
	
	# S[i] = F[i] + S[i-1]
	#      = sum(F[i-1]...F[i-N]) + S[i-1]
	#      = S[i-1] + S[i-1] - S[i-1-N]
	A[0][0] = 2
	A[0][N] = mo-1
	for i in range(1,N+1):
		A[i][i-1]=1
	
	# S[K]とS[K-1]を求める
	RetS = GetSum(A,K)
	PreS = GetSum(A,K-1)
	
	# F[K]=S[K]-S[K-1]
	RetF = (RetS - PreS + mo) % mo
	
	print "%d %d" % (RetF, RetS)