CodeForces 540B School Marks

http://codeforces.com/problemset/problem/540/B

 

School Marks

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

Little Vova studies programming in an elite school. Vova and his classmates are supposed to write n progress tests, for each test they will get a mark from 1 to p. Vova is very smart and he can write every test for any mark, but he doesn't want to stand out from the crowd too much. If the sum of his marks for all tests exceeds value x, then his classmates notice how smart he is and start distracting him asking to let them copy his homework. And if the median of his marks will be lower than y points (the definition of a median is given in the notes), then his mom will decide that he gets too many bad marks and forbid him to play computer games.

Vova has already wrote k tests and got marks a₁, ..., a_k. He doesn't want to get into the first or the second situation described above and now he needs to determine which marks he needs to get for the remaining tests. Help him do that.

Input

The first line contains 5 space-separated integers: n, k, p, x and y (1 ≤ n ≤ 999, n is odd, 0 ≤ k < n, 1 ≤ p ≤ 1000, n ≤ x ≤ n·p, 1 ≤ y ≤ p). Here n is the number of tests that Vova is planned to write, k is the number of tests he has already written, p is the maximum possible mark for a test, x is the maximum total number of points so that the classmates don't yet disturb Vova, y is the minimum median point so that mom still lets him play computer games.

The second line contains k space-separated integers: a₁, ..., a_k (1 ≤ a_i ≤ p) — the marks that Vova got for the tests he has already written.

Output

If Vova cannot achieve the desired result, print "-1".

Otherwise, print n - k space-separated integers — the marks that Vova should get for the remaining tests. If there are multiple possible solutions, print any of them.

Sample Input

Input

5 3 5 18 4
3 5 4

Output

4 1

Input

5 3 5 16 4
5 5 5

Output

-1

Hint

The median of sequence a₁, ..., a_n where n is odd (in this problem n is always odd) is the element staying on (n + 1) / 2 position in the sorted list of a_i.

In the first sample the sum of marks equals 3 + 5 + 4 + 4 + 1 = 17, what doesn't exceed 18, that means that Vova won't be disturbed by his classmates. And the median point of the sequence {1, 3, 4, 4, 5} equals to 4, that isn't less than 4, so his mom lets him play computer games.

Please note that you do not have to maximize the sum of marks or the median mark. Any of the answers: "4 2", "2 4", "5 1", "1 5", "4 1", "1 4" for the first test is correct.

In the second sample Vova got three '5' marks, so even if he gets two '1' marks, the sum of marks will be 17, that is more than the required value of 16. So, the answer to this test is "-1".

 

题目大意：

n道题，每道题的分值是1到p， Vova已经写了k道题， （1）Vova获得的总分数不能超过x，（2）中间的分数（即第（n+1)/2个分数（其中得到的所有的分数都是排过序的））不能超过y，问 Vova能否做到

不能做到输出-1，否则输出剩下n-k道题的得分

 

既然中间分数不能超过y，那么我们就假设中间分数就是y，先让 Vova的分数满足（2）， 那么接下来我们就要在保证中间数为y的前提下尽可能得使总分数小，

那么我们可以让中间数前面（中间数前面的得分必须比y要小，这样才能保证y是中间数）空着的地方得分为1，中间数后面空着的地方的得分为y，然后再判断总

分是否比x小

 

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<stack>
#include<algorithm>

using namespace std;
const int N = 1010;
typedef __int64 ll;

int main()
{
    int n, k, p, x, y, num;
    while(~scanf("%d%d%d%d%d", &n, &k, &p, &x, &y))
    {
        int m = 0, sum = 0;
        for(int i = 0 ;i < k ; i++)
        {
            scanf("%d", &num);
            sum += num;
            if(num < y)
                m++;//m统计比中间数y小的数的个数
        }
        if(m > n / 2)
        {
            printf("-1\n");
            continue;
        }
        int t1 = min(n - k, n / 2 - m);//中间数前面还有多少个空位来放1
        int t2 = n - k - t1;//中间数后面还有多少个空位来放y
        int sum2 = sum + t1 * 1 + t2 * y;//所得的总分数
        if(sum2 > x)
            printf("-1\n");
        else
        {
            for(int i = 0 ; i < t1 ; i++)
                printf("1 ");
            for(int i = 0 ; i < t2 ; i++)
                printf("%d ", y);
            printf("\n");
        }
    }
    return 0;
}
/*
9 7 2 14 1
2 2 2 1 1 2 2
*/