Problem: One very well-known internet resource site (let's call it X) has come up with a New Year adventure. Specifically, they decided to give ratings to all visitors.

There are n users on the site, for each user we know the rating value he wants to get as a New Year Present. We know that user i wants to get at least ai rating units as a present.

The X site is administered by very creative and thrifty people. On the one hand, they want to give distinct ratings and on the other hand, the total sum of the ratings in the present must be as small as possible.

Help site X cope with the challenging task of rating distribution. Find the optimal distribution.

Input

The first line contains integer n (1 ≤ n ≤ 3·105) — the number of users on the site. The next line contains integer sequence a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a sequence of integers b1, b2, ..., bn. Number bi means that user i gets bi of rating as a present. The printed sequence must meet the problem conditions. 

If there are multiple optimal solutions, print any of them.

Examples

Input

3
5 1 1


Output

5 1 2


Input

1
1000000000


Output

1000000000

Solution (in PYTHON):

Solution:
import sys
from collections import defaultdict, Counter
from math import sin, cos, asin, acos, tan, atan, pi

sys.setrecursionlimit(10 ** 6)

def pyes_no(condition, yes = "YES", no = "NO", none = "-1") :
  if condition == None:
    print (none)
  elif condition :
    print (yes)
  else :
    print (no)

def plist(a, s = ' ') :
  print (s.join(map(str, a)))

def rint() :
  return int(sys.stdin.readline())

def rstr() :
  return sys.stdin.readline().strip()


def rints() :
  return map(int, sys.stdin.readline().split())

def rfield(n, m = None) :
  if m == None :
    m = n
  
  field = []
  for i in xrange(n) :
    chars = sys.stdin.readline().strip()
    assert(len(chars) == m)
    field.append(chars)
  return field

def pfield(field, separator = '') :
  print ('\n'.join(map(lambda x: separator.join(x), field)))

def check_field_equal(field, i, j, value) :
  if i >= 0 and i < len(field) and j >= 0 and j < len(field[i]) :
    return value == field[i][j]
  return None 

def digits(x, p) :
  digits = []
  while x > 0 :
    digits.append(x % p)
    x //= p
  return digits[::-1]

def undigits(x, p) :
  value = 0
  for d in x :
    value *= p
    value += d
  return value

def modpower(a, n, mod) :
  r = a ** (n % 2)
  if n > 1 :
    r *= modpower(a, n // 2, mod) ** 2
  return r % mod

def gcd(a, b) :
  if a > b :
    a, b = b, a
  
  while a > 0 :
    a, b = b % a, a

  return b

def vector_distance(a, b) :
  diff = vector_diff(a, b)
  
  return scalar_product(diff, diff) ** 0.5

def vector_inverse(v) :
  r = [-x for x in v]

  return tuple(r)

def vector_diff(a, b) :
  return vector_sum(a, vector_inverse(b))

def vector_sum(a, b) :
  r = [c1 + c2 for c1, c2 in zip(a, b)]
    
  return tuple(r)

def scalar_product(a, b) :
  r = 0
  for c1, c2 in zip(a, b) :
    r += c1 * c2

  return r

def check_rectangle(points) :
  assert(len(points) == 4)

  A, B, C, D = points

  for A1, A2, A3, A4 in [
    (A, B, C, D),
    (A, C, B, D),
    (A, B, D, C),
    (A, C, D, B),
    (A, D, B, C),
    (A, D, C, B),
  ] :
    sides = (
      vector_diff(A1, A2),
      vector_diff(A2, A3),
      vector_diff(A3, A4),
      vector_diff(A4, A1),
    )
    if all(scalar_product(s1, s2) == 0 for s1, s2 in zip(sides, sides[1:])) :
       return True
  return False

def check_square(points) :
  if not check_rectangle(points) :
    return False
  A, B, C, D = points

  for A1, A2, A3, A4 in [
    (A, B, C, D),
    (A, C, B, D),
    (A, B, D, C),
    (A, C, D, B),
    (A, D, B, C),
    (A, D, C, B),
  ] :
    side_lengths = [
      (first[0] - next[0]) ** 2 + (first[1] - next[1]) ** 2 for first, next in zip([A1, A2, A3, A4], [A2, A3, A4, A1])
    ]
    if len(set(side_lengths)) == 1 :
      return True
    
  return False

def check_right(p) :
  # Check if there are same points
  for a, b in [
    (p[0], p[1]),
    (p[0], p[2]),
    (p[1], p[2]),
  ] :
    if a[0] == b[0] and a[1] == b[1] :
      return False

  a, b, c = p
  a, b, c = vector_diff(a, b), vector_diff(b, c), vector_diff(c, a)   

  return scalar_product(a, b) * scalar_product(a, c) * scalar_product(b, c) == 0

def modmatrixproduct(a, b, mod) :
  n, m1 = len(a), len(a[0])
  m2, k = len(b), len(b[0])

  assert(m1 == m2)
  m = m1

  r = [[0] * k for i in range(n)]
  for i in range(n) :
    for j in range(k) :
      for l in range(m) :
        r[i][j] += a[i][l] * b[l][j]
      r[i][j] %= mod
  return r

def modmatrixpower(a, n, mod) :
  magic = 2
  for m in [2, 3, 5, 7] :
    if n % m == 0 :
      magic = m
      break

  r = None
  if n < magic : 
    r = a
    n -= 1
  else :
    s = modmatrixpower(a, n // magic, mod)
    r = s
    for i in range(magic - 1) :
      r = modmatrixproduct(r, s, mod)

  for i in range(n % magic) : 
    r = modmatrixproduct(r, a, mod)
  
  return r

n = rint()
a = sorted(enumerate(rints()), key = lambda x: x[1])

ratings = [0] * n
previous = 0
for i, v in a :
  if previous >= v :
    v = previous + 1
  previous = v
  ratings[i] = v

plist(ratings)