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 ypoints (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, ..., ak. 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, nis 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, ..., ak (1 ≤ ai ≤ 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.

Examples

Input

5 3 5 18 4
3 5 4

Output

4 1

Input

5 3 5 16 4
5 5 5

Output

-1

Note

The median of sequence a1, ..., an 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 ai.

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".

題意：

給你n,k,p,x,y,五個數，分別表示小明要考n門課，現在已經考了k門課了，每門課的最高成績為p,但小明有兩個小要求。

1、小明考的n門課的總成績不能超過x;

2、小明的n門課的成績在排完序後，他的中位數要>=y。

問小明剩下的幾門課要考多少分，若有多種情況，輸出其中任意一種就行，如果不能滿足兩個條件，輸出-1.

//思路：

首先通過輸入的k門成績求出小明剩餘的幾門課，幾門是y以下的，幾門是y以上的。

求出這兩個數，下面的就簡單了，分別對它們賦值為1，y就行了（保證總分最小）。
C++版本一

 

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
#define N 10010
using namespace std;
int a[N];
int b[N];
int main()
{
	int n,k,p,x,y;
	while(scanf("%d%d%d%d%d",&n,&k,&p,&x,&y)!=EOF)
	{
		int i,j;
		int sum=0;
		for(i=1;i<=k;i++)
			scanf("%d",&a[i]),sum+=a[i];
		int sx=x-sum;
		for(i=k+1;i<=n;i++)
		{
			if(sx-(n-i)>=y)
				a[i]=y,sx-=y;
			else
				a[i]=1,sx-=1;
			b[i]=a[i];
		}
		sort(a+1,a+n+1);
		if(a[n/2+1]<y||sx<0)
			printf("-1\n");
		else
		{
			for(i=k+1;i<n;i++)
				printf("%d ",b[i]);
				printf("%d\n",b[n]);
		}
	}
	return 0;
}

 

C++版本二

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;
 
int n,p,k,x,y;
int a[maxn];
int ans[maxn];
 
int main()
{
//    freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
    int i,j;
    while (~scanf("%d%d%d%d%d",&n,&p,&k,&x,&y))
    {
        int s=0,l=0,r=0,num=0,cnt=0;
        for (i=0;i<p;i++)
        {
            scanf("%d",&a[i]);
            s+=a[i];
            if (a[i]<y) l++; //記錄比y小的數有多少
            else if (a[i]>=y) r++;//記錄大於等於y的數有多少
        }
        if (l>=n/2+1){  //如果比y小的數超過了一半，那麼中位數不可能達到y了
            printf("-1\n");
            continue;
        }
        int xx=n/2+1-r;
        while (xx>0){  //填充右邊，儘量填y使得和最小
            ans[cnt++]=y;
            xx--;s+=y;
        }
        xx=n-r-l-cnt;
        while (xx>0)    //填充左邊，填1使和儘量小
        {
            ans[cnt++]=1;
            xx--;s++;
        }
        if (s>x){   //和超過x輸出-1
            printf("-1\n");
            continue;
        }
        printf("%d",ans[0]);
        for (i=1;i<cnt;i++)
            printf(" %d",ans[i]);
        printf("\n");
    }
    return 0;
}