Nikephoros and Polycarpus play rock-paper-scissors. The loser gets pinched (not too severely!).

Let us remind you the rules of this game. Rock-paper-scissors is played by two players. In each round the players choose one of three items independently from each other. They show the items with their hands: a rock, scissors or paper. The winner is determined by the following rules: the rock beats the scissors, the scissors beat the paper and the paper beats the rock. If the players choose the same item, the round finishes with a draw.

Nikephoros and Polycarpus have played n rounds. In each round the winner gave the loser a friendly pinch and the loser ended up with a fresh and new red spot on his body. If the round finished in a draw, the players did nothing and just played on.

Nikephoros turned out to have worked out the following strategy: before the game began, he chose some sequence of items A = (a1, a2, ..., am), and then he cyclically showed the items from this sequence, starting from the first one. Cyclically means that Nikephoros shows signs in the following order: a1, a2, ..., am, a1, a2, ..., am, a1, ... and so on. Polycarpus had a similar strategy, only he had his own sequence of items B = (b1, b2, ..., bk).

Determine the number of red spots on both players after they've played n rounds of the game. You can consider that when the game began, the boys had no red spots on them.

Input

The first line contains integer n (1 ≤ n ≤ 2·109) — the number of the game's rounds.

The second line contains sequence A as a string of m characters and the third line contains sequence B as a string of k characters (1 ≤ m, k ≤ 1000). The given lines only contain characters "R", "S" and "P". Character "R" stands for the rock, character "S" represents the scissors and "P" represents the paper.

Output

Print two space-separated integers: the numbers of red spots Nikephoros and Polycarpus have.

Examples

Input

7
RPS
RSPP


Output

3 2

Input

5
RRRRRRRR
R


Output

0 0

Note

In the first sample the game went like this:

  * R - R. Draw. 
  * P - S. Nikephoros loses. 
  * S - P. Polycarpus loses. 
  * R - P. Nikephoros loses. 
  * P - R. Polycarpus loses. 
  * S - S. Draw. 
  * R - P. Nikephoros loses. 



Thus, in total Nikephoros has 3 losses (and 3 red spots), and Polycarpus only has 2.
def gen(str):
    def lol():
        while True:
            for ch in str:
                yield ch
    return lol
    
def nod(a, b):
  while a*b!=0:
    if a >= b:
      a = a%b
    else:
      b = b%a
  return a+b
  
def nok(a,b):
  return (a*b)//(nod(a,b))

n=int(raw_input())
A=raw_input()
B=raw_input()
ASH=0
BSH=0
i=0

lA=len(A)
lB=len(B)
NOK=nok(lA,lB)
#print NOK
pA=NOK//lA
pB=NOK//lB
sA=A*pA
sB=B*pB

NextA=gen(sA)()
NextB=gen(sB)()
A=NextA.next()
B=NextB.next()

if(n<=NOK): 
    while i<n:
        if A!=B:
            if A=="R" and B=="P":
                ASH+=1
            elif A=="R" and B=="S": 
                BSH+=1
            elif A=="S" and B=="R":
                ASH+=1
            elif A=="S" and B=="P": 
                BSH+=1
            elif A=="P" and B=="R": 
                BSH+=1
            else:
                ASH+=1
        i+=1
        A=NextA.next()
        B=NextB.next()
else:
    povt=n//NOK
    ost=n%NOK
    while i<NOK:
        if A!=B:
            if A=="R" and B=="P":
                ASH+=1
            elif A=="R" and B=="S": 
                BSH+=1
            elif A=="S" and B=="R":
                ASH+=1
            elif A=="S" and B=="P": 
                BSH+=1
            elif A=="P" and B=="R": 
                BSH+=1
            else:
                ASH+=1
        i+=1
        A=NextA.next()
        B=NextB.next()
    ASH*=povt
    BSH*=povt
    #print ASH, BSH
    
    NextA=gen(sA)()
    NextB=gen(sB)()
    A=NextA.next()
    B=NextB.next()
    i=0
    
    while i<ost:
        if A!=B:
            if A=="R" and B=="P":
                ASH+=1
            elif A=="R" and B=="S": 
                BSH+=1
            elif A=="S" and B=="R":
                ASH+=1
            elif A=="S" and B=="P": 
                BSH+=1
            elif A=="P" and B=="R": 
                BSH+=1
            else:
                ASH+=1
        i+=1
        A=NextA.next()
        B=NextB.next()
    
    
    
print ASH,BSH
