Ivan has an array consisting of n elements. Each of the elements is an integer from 1 to n.

Recently Ivan learned about permutations and their lexicographical order. Now he wants to change (replace) minimum number of elements in his array in such a way that his array becomes a permutation (i.e. each of the integers from 1 to n was encountered in his array exactly once). If there are multiple ways to do it he wants to find the lexicographically minimal permutation among them.

Thus minimizing the number of changes has the first priority, lexicographical minimizing has the second priority.

In order to determine which of the two permutations is lexicographically smaller, we compare their first elements. If they are equal — compare the second, and so on. If we have two permutations x and y, then x is lexicographically smaller if xi < yi, where i is the first index in which the permutations x and y differ.

Determine the array Ivan will obtain after performing all the changes.

Input

The first line contains an single integer n (2 ≤ n ≤ 200 000) — the number of elements in Ivan's array.

The second line contains a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ n) — the description of Ivan's array.

Output

In the first line print q — the minimum number of elements that need to be changed in Ivan's array in order to make his array a permutation. In the second line, print the lexicographically minimal permutation which can be obtained from array with q changes.

Examples

Input

4
3 2 2 3


Output

2
1 2 4 3 


Input

6
4 5 6 3 2 1


Output

0
4 5 6 3 2 1 


Input

10
6 8 4 6 7 1 6 3 4 5


Output

3
2 8 4 6 7 1 9 3 10 5 

Note

In the first example Ivan needs to replace number three in position 1 with number one, and number two in position 3 with number four. Then he will get a permutation [1, 2, 4, 3] with only two changed numbers — this permutation is lexicographically minimal among all suitable. 

In the second example Ivan does not need to change anything because his array already is a permutation.
import bisect
import decimal
from decimal import Decimal
import os
from collections import Counter
import bisect
from collections import defaultdict
import math
import random
import heapq 
from math import sqrt
import sys
from functools import reduce, cmp_to_key
from collections import deque
import threading
from itertools import combinations
from io import BytesIO, IOBase
from itertools import accumulate

# sys.setrecursionlimit(200000)
# mod = 10**9+7
# mod = 998244353
decimal.getcontext().prec = 46

def primeFactors(n): 
    prime = set()
    while n % 2 == 0: 
        prime.add(2)
        n = n//2
    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0: 
            prime.add(i)
            n = n//i 
    if n > 2: 
        prime.add(n)
    return list(prime)
    
def getFactors(n) : 
    factors = []
    i = 1
    while i <= math.sqrt(n): 
        if (n % i == 0) : 
            if (n // i == i) : 
                factors.append(i)
            else : 
                factors.append(i)
                factors.append(n//i)
        i = i + 1
    return factors
    
def SieveOfEratosthenes(n): 
    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
    num = []
    for p in range(2, n+1): 
        if prime[p]: 
            num.append(p)
    return num

def lcm(a,b):
    return (a*b)//math.gcd(a,b)

def sort_dict(key_value):
    return sorted(key_value.items(), key = lambda kv:(kv[1], kv[0]))

def list_input():
    return list(map(int,input().split()))
    
def num_input():
    return map(int,input().split())
    
def string_list():
    return list(input())
    
def decimalToBinary(n):  
    return bin(n).replace("0b", "")  
    
def binaryToDecimal(n): 
    return int(n,2) 
    
def DFS(n,s,adj):              
    visited = [False for i in range(n)]  
    stack = [] 
    stack.append(s)  
  
    while (len(stack)):  
        s = stack[-1]  
        stack.pop() 
  
        if (not visited[s]):  
            visited[s] = True 
 
        for node in adj[s]:  
            if (not visited[node]):  
                stack.append(node)  
        
def solve():
    n = int(input())
    arr = list_input()
    nums = {}
    not_visited = {}
    for i in range(n):
        not_visited[i+1] = 1
    for i in range(n):
        try:
            nums[arr[i]] += 1
        except KeyError:
            nums[arr[i]] = 1
            del not_visited[arr[i]]
            
    key = list(nums.keys())
    
    if len(nums.keys()) == n:
        print(0)
        print(*arr)
        return
    
    not_visited = list(not_visited.keys())
    visited = {}
    op = 0
    indx = 0
    for i in not_visited:
        while indx < n:
            if arr[indx] > i:
                if nums[arr[indx]] > 1:
                    op += 1
                    nums[arr[indx]] -= 1
                    arr[indx] = i
                    indx += 1
                    break
                else:
                    visited[arr[indx]] = 1
                    indx += 1
            elif arr[indx] < i:
                try:
                    visited[arr[indx]]
                    if nums[arr[indx]] > 1:
                        op += 1
                        nums[arr[indx]] -= 1
                        arr[indx] = i
                        indx += 1
                        break
                except KeyError:
                    visited[arr[indx]] = 1
                    indx += 1
            
    print(op)
    print(*arr)
        

#t = int(input())
t = 1
for _ in range(t):
    solve()


    
    
    
    
