time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Dasha logged into the system and began to solve problems. One of them is as follows:

Given two sequences a and b of length n each you need to write a sequence c of length n, the i-th element of which is calculated as follows: c__i = b__i - a__i.

About sequences a and b we know that their elements are in the range from l to r. More formally, elements satisfy the following conditions: l ≤ a__i ≤ r and l ≤ b__i ≤ r. About sequence c we know that all its elements are distinct.

Dasha wrote a solution to that problem quickly, but checking her work on the standard test was not so easy. Due to an error in the test system only the sequence a and the compressed sequence of the sequence c were known from that test.

Let's give the definition to a compressed sequence. A compressed sequence of sequence c of length n is a sequence p of length n, so that p__i equals to the number of integers which are less than or equal to c__i in the sequence c. For example, for the sequence c = [250, 200, 300, 100, 50] the compressed sequence will be p = [4, 3, 5, 2, 1]. Pay attention that in c all integers are distinct. Consequently, the compressed sequence contains all integers from 1 to n inclusively.

Help Dasha to find any sequence b for which the calculated compressed sequence of sequence c is correct.

Input

The first line contains three integers n, l, r (1 ≤ n ≤ 105, 1 ≤ l ≤ r ≤ 109) — the length of the sequence and boundaries of the segment where the elements of sequences a and b are.

The next line contains n integers a_1,  _a_2,  …,  _a__n (l ≤ a__i ≤ r) — the elements of the sequence a.

The next line contains n distinct integers p_1,  _p_2,  …,  _p__n (1 ≤ p__i ≤ n) — the compressed sequence of the sequence c.

Output

If there is no the suitable sequence b, then in the only line print "-1".

Otherwise, in the only line print n integers — the elements of any suitable sequence b.

Examples

input

5 1 5
1 1 1 1 1
3 1 5 4 2

output

3 1 5 4 2

input

4 2 9
3 4 8 9
3 2 1 4

output

2 2 2 9

input

6 1 5
1 1 1 1 1 1
2 3 5 4 1 6

output

-1

给一个数列a，给一个数列p，已知数列p是数列c按个元素大小编号1~n后的所谓的压缩数列，求任意一个符合条件的数列b=a+c（条件：数列a、b都在某个范围内）；

思路：

例如对于第二组样例，

3 4 8 9

根据 3 2 1 4的大小重排后得到

8 4 3 9 (a)

1 2 3 4 (p)

相对应的，不妨假设数列b对应的第一位为下界2，那么可知数列c的对应第一位为-6

8 4 3 9 (a)

1 2 3 4 (p)

-6          (c)

2           (b)

再看第二位，不妨假设b还是下界，则c为-2，-2 > -6，符合p给定的大小顺序，则可以有：

8 4 3 9 (a)

1 2 3 4 (p)

-6 -2        (c)

2  2      (b)

再到第三位，不妨设b还是下界，则c为-1，依然符合p的大小顺序

8  4  3  9 (a)

1  2  3  4 (p)

-6 -2 -1     (c)

2  2  2     (b)

再到第四位，设b为下界，则c为-7，这时候发现不符合顺序，那么就让c等于前一位+1

8   4   3   9  (a)

1   2   3   4  (p)

-6  -2  -1  0  (c)

2   2   2   9   (b)

这时候，得到的b为9，不大于上界，符合条件，因此我们得到了一个符合条件的b：

2   2   2   9   (b)

最后我们在将数列b排回3 2 1 4的顺序即可（这步很重要……在这个样例里重新排序之后没有变化，但其他样例就不一定了……）

3   2   1   4  (p)

2   2   2   9   (b)

感觉自己写的代码一点都不优雅……（羞耻……

#include
int main()
{
int n,l,r,a[+],temp_a[+],p[+],c[+];
scanf("%d %d %d",&n,&l,&r);
for(int i=;i<=n;i++) scanf("%d",&temp_a[i]);
for(int i=;i<=n;i++) scanf("%d",&p[i]);
for(int i=;i<=n;i++) a[ p[i] ]=temp_a[i];

 //for(int i=1;i<=n;i++) printf("%d ",a\[i\]);printf("\\n");

 int now=l-a\[\]-;  
 for(int i=;i<=n;i++)  
 {  
     if(l-a\[i\] > now) c\[i\]=(now=l-a\[i\]);  
     else c\[i\]=(now+=);  
     if( c\[i\]+a\[i\] > r ){  
         printf("-1\\n");  
         return ;  
     }  
 }

 for(int i=;i<=n;i++){  
     if(i!=) printf(" ");  
     printf("%d",c\[p\[i\]\]+a\[p\[i\]\]);  
 }  
 printf("\\n");  
 return ;  

}