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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.

题意：给你一组长度为n的序列，里面由1~n中的数字组成（有可能有重复）。要求是我们要把这个序列弄成字典序最小的n的一个全排列。 思路：首先我们把没有出现过的数字放到一个集合中。 两种情况： ①如果这个数字出现超过一次并且字典序大于集合中的第一个元素，就替换掉。 ②如果这个数字出现超过一次并且字典序小于集合中的第一个元素，就不替换，等到下一次出现的时候替换掉。 代码如下：

#include
   
    #define ll long longusing namespace std;const int maxx=2e5+100;int a[maxx];int c[maxx];int vis[maxx];int n;int main(){
   	scanf("%d",&n);	for(int i=1;i<=n;i++)	{
   		scanf("%d",&a[i]);		c[a[i]]++;	}	set
    
      p;	for(int i=1;i<=n;i++)	{
   		if(c[i]==0) p.insert(i);	}	int ans=0;	for(int i=1;i<=n;i++)	{
   		if(c[a[i]]>1)		{
   			if(*p.begin()
    
   

努力加油a啊，(^o)/~

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