Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of s characters. The first participant types one character in v1 milliseconds and has ping t1 milliseconds. The second participant types one character in v2 milliseconds and has ping t2 milliseconds.

If connection ping (delay) is t milliseconds, the competition passes for a participant as follows: 

  1. Exactly after t milliseconds after the start of the competition the participant receives the text to be entered. 
  2. Right after that he starts to type it. 
  3. Exactly t milliseconds after he ends typing all the text, the site receives information about it. 



The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.

Given the length of the text and the information about participants, determine the result of the game.

Input

The first line contains five integers s, v1, v2, t1, t2 (1 ≤ s, v1, v2, t1, t2 ≤ 1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.

Output

If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".

Examples

Input

5 1 2 1 2


Output

First


Input

3 3 1 1 1


Output

Second


Input

4 5 3 1 5


Output

Friendship

Note

In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.

In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.

In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw.
# -*- coding: utf-8 -*-
import sys
import math
import os
import itertools
import string
import heapq
import _collections
from collections import Counter
from collections import defaultdict
from functools import lru_cache
import bisect
import re
import queue
from decimal import *


class Scanner():
    @staticmethod
    def int():
        return int(sys.stdin.readline().rstrip())

    @staticmethod
    def string():
        return sys.stdin.readline().rstrip()

    @staticmethod
    def map_int():
        return [int(x) for x in Scanner.string().split()]

    @staticmethod
    def string_list(n):
        return [input() for i in range(n)]

    @staticmethod
    def int_list_list(n):
        return [Scanner.map_int() for i in range(n)]

    @staticmethod
    def int_cols_list(n):
        return [int(input()) for i in range(n)]


class Math():
    @staticmethod
    def gcd(a, b):
        if b == 0:
            return a
        return Math.gcd(b, a % b)

    @staticmethod
    def lcm(a, b):
        return (a * b) // Math.gcd(a, b)

    @staticmethod
    def divisor(n):
        res = []
        i = 1
        for i in range(1, int(n ** 0.5) + 1):
            if n % i == 0:
                res.append(i)
                if i != n // i:
                    res.append(n // i)
        return res

    @staticmethod
    def round_up(a, b):
        return -(-a // b)

    @staticmethod
    def is_prime(n):
        if n < 2:
            return False
        if n == 2:
            return True
        if n % 2 == 0:
            return False
        d = int(n ** 0.5) + 1
        for i in range(3, d + 1, 2):
            if n % i == 0:
                return False
        return True


class PriorityQueue:
    def __init__(self, l=[]):
        self.q = l
        heapq.heapify(self.q)
        return

    def push(self, n):
        heapq.heappush(self.q, n)
        return

    def pop(self):
        return heapq.heappop(self.q)


def pop_count(x):
    x = x - ((x >> 1) & 0x5555555555555555)
    x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
    x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
    x = x + (x >> 8)
    x = x + (x >> 16)
    x = x + (x >> 32)
    return x & 0x0000007f


MOD = int(1e09) + 7
INF = int(1e15)


def main():
    # sys.stdin = open("sample.txt")
    s, v1, v2, t1, t2 = Scanner.map_int()
    a = s * v1 + t1 * 2
    b = s * v2 + t2 * 2
    if a == b:
        print('Friendship')
    elif a < b:
        print('First')
    else:
        print('Second')


if __name__ == "__main__":
    main()
