Codeforces Round #436 (Div. 2)D. Make a Permutation! codeforces-864D. Make a Permutation!

D. Make a Permutation!

time limit per test

 2 seconds

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

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

43 2 2 3

output

21 2 4 3 

input

64 5 6 3 2 1

output

04 5 6 3 2 1 

input

106 8 4 6 7 1 6 3 4 5

output

32 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：给你一段从1到n的序列，其中可能存在重复的元素，现在让你将那些1-n中没有出现在序列中的数字拿过去替换重复的元素。然后最后的序列中只有1-n，并且所有的元素只出现了一次。现在要求你输出最少的替换次数，使得最后的序列的字典序最小。（第一排序为替换次数，第二排序为字典序）

思路：我们先在输入时记录出现的次数，这里开一个vis数组记录对应下标的出现次数。然后再去从已知的序列的开始判断，寻找那些出现了不止一次的数字，然后判断此时此刻应该补充的最小数字（这里开一个循环寻找那个最小的没有出现的数字然后记录此时的数字），如果此时需要替换的数字比原始的数字还要小那么我们就去替换这个数字，否则就是不管继续进行下一步操作。替换之后注意一些后续的标记就行了。所以这题就是这么愉快的结束了。

代码：

[cpp] view plain copy

1. #include<bits/stdc++.h>  
2. using namespace std;  
3.   
4. int num[200005];  
5. int vis[200005];  
6. int have[200005];  
7.   
8. int main()  
9. {  
10.     int N;;  
11.     while(~scanf("%d",&N))  
12.     {  
13.         memset(vis,0,sizeof vis);  
14.         memset(have,0,sizeof have);  
15.         int cnum=1;  
16.         int cnt=0;  
17.         for(int i=0;i<N;i++)  
18.         {  
19.             scanf("%d",&num[i]);  
20.             vis[num[i]]++;  
21.         }  
22.         for(int i=0;i<N;i++)  
23.         {  
24.             if(vis[num[i]]>1)  
25.             {  
26.                 while(vis[cnum])  
27.                     cnum++;  
28.                 //cout<<cnum<<endl;  
29.                 if(cnum>=num[i]&&!have[num[i]])  
30.                 {  
31.                     have[num[i]]++;  
32.                     continue;  
33.                 }  
34.                 cnt++;  
35.                 vis[num[i]]--;  
36.                 vis[cnum]++;  
37.                 num[i]=cnum;  
38.             }  
39.         }  
40.         printf("%d\n%d",cnt,num[0]);  
41.         for(int i=1;i<N;i++)  
42.             printf(" %d",num[i]);  
43.         printf("\n");  
44.     }  
45.     return 0;  
46. }