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 java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

 
public class Try
{
    
    static class InputReader {
 
		private final InputStream stream;
		private final byte[] buf = new byte[8192];
		private int curChar, snumChars;
 
		public InputReader(InputStream st) {
			this.stream = st;
		}
 
		public int read() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}
 
		public int ni() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public long nl() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public int[] nia(int n) {
			int a[] = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = ni();
			}
			return a;
		}
 
		public String rs() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
 
		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}
 
		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
 
		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
 
	}
	

	
	
	
    static long mod=1000000007;
    static BigInteger bigInteger = new BigInteger("1000000007");
    static int n = (int)1e6;
    static boolean[] prime;
    static ArrayList<Integer> as;
    static void sieve() {
			Arrays.fill(prime	, true);
			prime[0] = prime[1] = false;
			for(int i = 2 ; i * i <= n ; ++i) {
				if(prime[i]) {
					for(int k = i * i; k< n ; k+=i) {
						prime[k] = false;
					}
				}
			}
			
		}
		static PrintWriter w = new PrintWriter(System.out);
 
    public static void main(String[] args)
    {
        
        InputReader sc = new InputReader(System.in);
        //PrintWriter w = new PrintWriter(System.out);
    
       /* 
        prime  = new boolean[n + 1];
		sieve();
		prime[1] = false;
		*/
		/*
		as = new ArrayList<>();
		for(int i=2;i<=1000000;i++)
		{
		    if(prime[i])
		    as.add(i);
		}
		*/
		
       /* 
		
	long a = sc.nl();
	 
            BigInteger ans = new BigInteger("1");
            for (long i = 1; i < Math.sqrt(a); i++) {
                if (a % i == 0) {
                    if (a / i == i) {
                        ans = ans.multiply(BigInteger.valueOf(phi(i)));
                        
                    } else {
                        ans = ans.multiply(BigInteger.valueOf(phi(i)));
                        ans = ans.multiply(BigInteger.valueOf(phi(a / i)));
 
                        
                    }
 
 
                }
            }
            w.println(ans.mod(bigInteger));
            */
          
         // MergeSort ob = new MergeSort();
       // ob.sort(arr, 0, arr.length-1);
     
   
  
   
 
         int x = sc.ni();
         int []a = sc.nia(x);
         int []arr = new int[200001];
         for(int i=0;i<x;i++)
         {
             arr[a[i]]++;
             
         }
      int res = 0;
      TreeSet<Integer> ts = new TreeSet<>();
      int []numarr = new int[x+1];
      int []free = new int[x+1];
      int c = 0;
      int maxfree = 0;
      for(int i=1;i<200001;i++)
      {
          if(i <= x)
          {
              if(arr[i] == 0)
              {
                  ts.add(i);
                  free[c] = i;
                  c++;
                  maxfree++;
              }
          }
      }
      for(int i=1;i<200001;i++)
      {
          if(arr[i] > 0)
              res += (arr[i] - 1);
      }
     
      if(res!=0)
      {
          int d = 0;
         
          for(int i=0;i<x;i++)
          {
              if(arr[a[i]] > 1)
              {
                  if(a[i] > free[d] || numarr[a[i]] > 0)
                  {
                      arr[a[i]]--;
                      a[i] = free[d];
                      d++;
                      
                  }
                  else
                      numarr[a[i]]++;
                
              }
             
          }
      }
      w.println(res);
      for(int i=0;i<x;i++)
      {
          w.print(a[i] + " ");
      }
      
      w.close();
        
    
    
    
    
    
   
  
    
    }
     
     public static long modMultiply(long one, long two) {
 
		return (one % mod * two % mod) % mod;
	}
    static long fx(int m)
    {
        long re = 0;
        for(int i=1;i<=m;i++)
        {
            re += (long) (i / gcd(i,m));
        }
        return re;
    }
     static long gcd(long a, long b)
    {
	 if (a == 0)
	  return b;
				         
	return gcd(b%a, a);
    }
    
   
    
     static long phi(long nx)
    {
        // Initialize result as n
        double result = nx; 
        
     
        // Consider all prime factors of n and for 
        // every prime factor p, multiply result
        // with (1 - 1/p)
        for (int p = 0; as.get(p) * as.get(p) <= nx; ++p)
        {
            // Check if p is a prime factor.
            if (nx % as.get(p) == 0)
            {
                // If yes, then update n and result
                while (nx % as.get(p) == 0)
                    nx /= as.get(p);
                result *= (1.0 - (1.0 / (double) as.get(p)));
            }
        }
     
        // If n has a prime factor greater than sqrt(n)
        // (There can be at-most one such prime factor)
        if (nx > 1)
            result *= (1.0 - (1.0 / (double) nx));
     
        return (long)result;
    
        
        //return(phi((long)result,k-1));
        
    }
    
    public static int primeFactors(int n)
    {
        int sum = 0;
        // Print the number of 2s that divide n
        while (n%2==0)
        {
            sum = 1;
            //System.out.print(2 + " ");
            n /= 2;
        }
 
        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            // While i divides n, print i and divide n
            while (n%i == 0)
            {
            
               // System.out.print(i + " ");
                n /= i;
            }
            sum++;
        }
 
        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n > 2)
            sum++;
            return sum;
    }
    static int digitsum(int x)
    
    {    
 
        int sum = 0;
        while(x > 0)
        {
            int temp = x % 10;
            sum += temp;
            x /= 10;
        }
        return sum;
        
    }
    static int countDivisors(int n)
    {
        int cnt = 0;
        for (int i = 1; i*i <=n; i++)
        {
             if (n % i == 0 && i<=1000000)
             {
                // If divisors are equal,
                // count only one
                if (n / i == i)
                   cnt++;
      
                else // Otherwise count both
                    cnt = cnt + 2;
             }
         }
        return cnt;
    }
   
    static boolean isprime(int n)
    {
    
    if(n == 2)
        return true;
    if(n == 3)
        return true;
    if(n % 2 == 0)
        return false;
    if(n % 3 == 0)
        return false;
 
    int i = 5;
   int w = 2;
 
    while(i * i <= n)
    {
        if(n % i == 0)
            return false;
    
 
        i += w;
        w = 6 - w;
    }
 
    return true;
    }
    
    static long log2(long value) {
    return Long.SIZE-Long.numberOfLeadingZeros(value);
}
 
 
 
 
    
    
    static boolean binarysearch(int []arr,int p,int n)
    {
        //ArrayList<Integer> as = new ArrayList<>();
        //as.addAll(0,at);
        //Collections.sort(as);
        boolean re = false;
        int st = 0;
        int end = n-1;
        
        
        
        while(st <= end)
        { 
            int mid = st + (end-st)/2;
        
            if(p > as.get(mid))
            {
                st = mid+1;
                
            }
            else if(p < as.get(mid))
            {
                end = mid-1;
                
            }
            else if(p == as.get(mid))
            {
            
            re = true;
            break;
            }
           
            
            
        }
         return re;
        
    }
    
   
    static class Student
    {
        
        int position;
        int value;
        Student(int position,int value)
        {
            this.position = position;
            this.value = value;
            
            
        }
        
    }
    /* Java program for Merge Sort */
static class MergeSort
{
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];
 
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
    void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
 
    /* A utility function to print array of size n */
    
 
   
}
    

    
    
    
    public static int ip(String s){
		return Integer.parseInt(s);
	}
	 public static String ips(int s){
		return Integer.toString(s);
	}
	
    
					 
}
 
