題干：

Eugeny has?n?cards, each of them has exactly one integer written on it. Eugeny wants to exchange some cards with Nikolay so that the number of even integers on his cards would equal the number of odd integers, and that all these numbers would be?distinct.

Nikolay has?m?cards, distinct numbers from?1?to?m?are written on them, one per card. It means that Nikolay has exactly one card with number?1, exactly one card with number?2?and so on.

A single exchange is a process in which Eugeny gives one card to Nikolay and takes another one from those Nikolay has. Your task is to find the minimum number of card exchanges and determine which cards Eugeny should exchange.

Input

The first line contains two integers?n?and?m?(2?≤?n?≤?2·105,?1?≤?m?≤?109)?— the number of cards Eugeny has and the number of cards Nikolay has. It is guaranteed that?n?is even.

The second line contains a sequence of?n?positive integers?a1,?a2,?...,?an?(1?≤?ai?≤?109)?— the numbers on Eugeny's cards.

Output

If there is no answer, print?-1.

Otherwise, in the first line print the minimum number of exchanges. In the second line print?n?integers?— Eugeny's cards after all the exchanges with Nikolay. The order of cards should coincide with the card's order in the input data. If the?i-th card wasn't exchanged then the?i-th number should coincide with the number from the input data. Otherwise, it is considered that this card was exchanged, and the?i-th number should be equal to the number on the card it was exchanged to.

If there are multiple answers, it is allowed to print any of them.

Examples

Input

6 2 5 6 7 9 4 5

Output

1 5 6 7 9 4 2

Input

8 6 7 7 7 7 8 8 8 8

Output

6 7 2 4 6 8 1 3 5

Input

4 1 4 2 1 10

Output

-1

題目大意：

尤金有?n?張卡片。每張卡片上都有一個整數。 尤金希望與尼古拉交換一些卡片，使得他的偶數卡片上的數量等于奇數卡片的數量，并且所有這些數字不同的。

尼古拉有?m?張卡，上面的數字為?1?到?m?，也就是每種數字的卡各一張。

每次兩人可以交換一張卡，問最少交換幾次可以滿足尤金的要求。

Input

第一行包含兩個整數?n?和?m?(2?≤?n?≤?2·105,?1?≤?m?≤?109)?— 分別表示兩人擁有的卡片數。 保證?n?是偶數。

第二行包含一個由?n?個正整數組成的數列?a1,?a2,?...,?an?(1?≤?ai?≤?109)?— 表示尤金的卡。

Output

如果沒有答案，輸出?-1.

否則，在第一行輸出最小需要換的卡片數。在第二行輸出?n?個整數?— 交換后尤金的卡牌。 卡牌的順序要和讀入時一樣，如果有多種方案，輸出任意一種。

解題報告：

先預處理出來在<=m的前提下有多少奇數和偶數可以使用，然后先保證能不變的先不變（打個ok標記），剩下的貪心填。注意一下貪心的順序，并且時刻注意填的數要<=m這個限制就好。

注意一定要先保證能不變的就不變，不然的話你不確定后面已經有多少奇數偶數了，那么就不能確定這一位優先填奇數還是偶數，所有有可能操作次數不是最少的。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 6e5 + 5; ll n,m,ji,ou,resji,resou,needji,needou; ll a[MAX]; set<ll> sji,sou,ans; bool vis[MAX],ok[MAX]; vector<int> o,j; int main() {cin>>n>>m;if(m&1) resou=m/2,resji=resou+1;else resou=resji=m/2;for(int i = 1; i<=n; i++) { scanf("%lld",a+i);if(a[i] < MAX) vis[a[i]]=1;if(a[i]&1) {ji++;if(a[i] <= m && sji.find(a[i]) == sji.end()) resji--;sji.insert(a[i]);}else {ou++;if(a[i] <= m && sou.find(a[i]) == sou.end()) resou--;sou.insert(a[i]);}}ll bij = n/2-sji.size();//必須要去掉的奇數，準確的說，是還需要換上的奇數，還缺少的奇數ll bio = n/2-sou.size();//必須要去掉的偶數if(resji < bij || resou < bio) return 0,puts("-1");ll tmp = max(bij,0LL)+max(bio,0LL);printf("%lld\n",tmp);for(int i = 2; i<=min(m,1LL*MAX-1); i+=2) {if(vis[i]) continue;o.pb(i);}for(int i = 1; i<=min(m,1LL*MAX-1); i+=2) {if(vis[i]) continue;j.pb(i);}ll curji=0,curou=0;for(int i = 1; i<=n; i++) {if(ans.count(a[i])) continue;ans.insert(a[i]);if((a[i]&1) && curji < n/2) curji++,ok[i]=1;else if((a[i]&1)==0 && curou<n/2)curou++,ok[i]=1;}for(int i = 1; i<=n; i++) {if(ok[i]) {continue;}if(a[i]&1) {//如果是奇數if(ans.count(a[i])) {if(curji < n/2) a[i]=j.back(),j.pop_back(),curji++;else if(curou < n/2) a[i] = o.back(),o.pop_back(),curou++;}else {if(curji < n/2) ans.insert(a[i]),curji++;else a[i] = o.back(),o.pop_back(),curou++;}}else {if(ans.count(a[i])) {if(curji < n/2) a[i]=j.back(),j.pop_back(),curji++;else if(curou < n/2) a[i] = o.back(),o.pop_back(),curou++;}else {if(curou < n/2) ans.insert(a[i]),curou++;else a[i] = j.back(),j.pop_back(),curji++;}}}for(int i = 1; i<=n; i++) printf("%lld ",a[i]);return 0 ; }

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