Vus the Cossack has n real numbers a_i. It is known that the sum of all numbers is equal to 0. He wants to choose a sequence b the size of which is n such that the sum of all numbers is 0 and each b_i is either ⌊ a_i ⌋ or ⌈ a_i ⌉. In other words, b_i equals a_i rounded up or down. It is not necessary to round to the nearest integer.

For example, if a = [4.58413, 1.22491, -2.10517, -3.70387], then b can be equal, for example, to [4, 2, -2, -4]. 

Note that if a_i is an integer, then there is no difference between ⌊ a_i ⌋ and ⌈ a_i ⌉, b_i will always be equal to a_i.

Help Vus the Cossack find such sequence!

Input

The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of numbers.

Each of the next n lines contains one real number a_i (|a_i| < 10^5). It is guaranteed that each a_i has exactly 5 digits after the decimal point. It is guaranteed that the sum of all the numbers is equal to 0.

Output

In each of the next n lines, print one integer b_i. For each i, |a_i-b_i|<1 must be met.

If there are multiple answers, print any.

Examples

Input


4
4.58413
1.22491
-2.10517
-3.70387


Output


4
2
-2
-4


Input


5
-6.32509
3.30066
-0.93878
2.00000
1.96321


Output


-6
3
-1
2
2

Note

The first example is explained in the legend.

In the second example, we can round the first and fifth numbers up, and the second and third numbers down. We can round the fourth number neither up, nor down.
from __future__ import division, print_function


def main():
    # Template 1.0
    import sys, re, math
    from collections import deque, defaultdict, Counter, OrderedDict
    from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, floor
    from heapq import heappush, heappop, heapify, nlargest, nsmallest
    def STR(): return list(input())

    def INT(): return int(input())

    def MAP(): return map(int, input().split())

    def LIST(): return list(map(int, input().split()))

    def list2d(a, b, c): return [[c] * b for i in range(a)]

    def sortListWithIndex(listOfTuples, idx):   return (sorted(listOfTuples, key=lambda x: x[idx]))

    def sortDictWithVal(passedDic):
        temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))
        toret = {}
        for tup in temp:
            toret[tup[0]] = tup[1]
        return toret

    def sortDictWithKey(passedDic):
        return dict(OrderedDict(sorted(passedDic.items())))

    INF = float('inf')
    mod = 10 ** 9 + 7


    n = INT()
    a = []
    checkForInt = [0]*(n+1)
    for _ in range(n):
        temp = float(input())
        a.append(int(floor(temp)))
        if(temp.is_integer()):
            checkForInt[_] = 1
    sum1 = sum(a)
    i = 0
    # print(checkForInt)
    while(sum1!=0):
        if(checkForInt[i]==1):
            i+=1
            continue
        a[i]+=1
        sum1+=1

        i+=1
    for el in a:
        print(el)
######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip

    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192


class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')

# Cout implemented in Python
import sys


class ostream:
    def __lshift__(self, a):
        sys.stdout.write(str(a))
        return self


cout = ostream()
endl = '\n'


# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero=0):
    conv = ord if py2 else lambda x: x
    A = [];
    numb = zero;
    sign = 1;
    i = 0;
    s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0'[0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-'[0]:
                sign = -1
            elif s[i] != b'\r'[0]:
                A.append(sign * numb)
                numb = zero;
                sign = 1
            i += 1
    except:
        pass
    if s and s[-1] >= b'0'[0]:
        A.append(sign * numb)
    return A


if __name__ == "__main__":
    main()