Misha and Vanya have played several table tennis sets. Each set consists of several serves, each serve is won by one of the players, he receives one point and the loser receives nothing. Once one of the players scores exactly k points, the score is reset and a new set begins.

Across all the sets Misha scored a points in total, and Vanya scored b points. Given this information, determine the maximum number of sets they could have played, or that the situation is impossible.

Note that the game consisted of several complete sets.

Input

The first line contains three space-separated integers k, a and b (1 ≤ k ≤ 109, 0 ≤ a, b ≤ 109, a + b > 0).

Output

If the situation is impossible, print a single number -1. Otherwise, print the maximum possible number of sets.

Examples

Input

11 11 5


Output

1


Input

11 2 3


Output

-1

Note

Note that the rules of the game in this problem differ from the real table tennis game, for example, the rule of "balance" (the winning player has to be at least two points ahead to win a set) has no power within the present problem.
#Fast IO Region
import os
import sys
from io import BytesIO,IOBase
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# Get out of main functoin
def main():
    pass
# decimal to binary
def binary(n):
    return (bin(n).replace("0b", ""))
# binary to decimal
def decimal(s):
    return (int(s, 2))
# power of a number base 2
def pow2(n):
    p = 0
    while n > 1:
        n //= 2
        p += 1
    return (p)
# if  number is prime in √n time
def isPrime(n):
    if (n == 1):
        return (False)
    else:
        root = int(n ** 0.5)
        root += 1
        for i in range(2, root):
            if (n % i == 0):
                return (False)
        return (True)
# list to string ,no spaces
def lts(l):
    s = ''.join(map(str, l))
    return s
# String to list
def stl(s):
    # for each character in string to list with no spaces -->
    l = list(s)
    # for space in string  -->
    # l=list(s.split(" "))
    return l
# Returns list of numbers with a particular sum
def sq(a, target, arr=[]):
    s = sum(arr)
    if (s == target):
        return arr
    if (s >= target):
        return
    for i in range(len(a)):
        n = a[i]
        remaining = a[i + 1:]
        ans = sq(remaining, target, arr + [n])
        if (ans):
            return ans
# Sieve for prime numbers in a range
def SieveOfEratosthenes(n):
    cnt = 0
    prime = [True for i in range(n + 1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1
    for p in range(2, n + 1):
        if prime[p]:
            cnt += 1
            # print(p)
    return (cnt)
# for positive integerse only
def nCr(n, r):
    f = math.factorial
    return f(n) // f(r) // f(n - r)
# 1000000007
mod = int(1e9) + 7
def ssinp(): return input()
# s=input()
def iinp(): return int(input())
# n=int(input())
def nninp(): return map(int,input().split())
#a,b,c=map(int,input().split())
def llinp(): return list(map(int,input().split()))
# a=list(map(int,input().split()))
def p(xyz): print(xyz)
def p2(a,b): print(a,b)
import math
#import random
#sys.setrecursionlimit(300000)
#from fractions import Fraction
#from collections import OrderedDict
#from collections import deque
########################                         mat=[[0 for i in range(n)] for j in range(m)]                      ########################
########################                         list.sort(key=lambda x:x[1]) for sorting a list according to second element in sublist            ########################
########################                         Speed: STRING < LIST < SET,DICTIONARY                      ########################
########################                         from collections import deque                              ########################
########################                         ASCII of A-Z= 65-90                                        ########################
########################                         ASCII of a-z= 97-122                                       ########################
########################                         d1.setdefault(key, []).append(value)                       ########################
#for __ in range(iinp()):
k,a,b=nninp()
x=a//k+b//k
if x==0:
    p(-1)
elif (a%k!=0 and b<k):
    p(-1)
elif (b%k!=0 and a<k):
    p(-1)
else:
    p(x)




""" Stuff you should look for
    1.int overflow, array bounds
    2.special cases (n=1?)
    3. do something instead of nothing and stay organized
    4. WRITE STUFF DOWN (MAIN)
	5. DON'T GET STUCK ON ONE APPROACH"""








