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: ci = bi - ai.
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 ≤ ai ≤ r and l ≤ bi ≤ 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 pi equals to the number of integers which are less than or equal to ci 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 a1, a2, ..., an (l ≤ ai ≤ r) — the elements of the sequence a.
The next line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ 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……
1 #include<cstdio>
2 int main()
3 {
4 int n,l,r,a[100000+5],temp_a[100000+5],p[100000+5],c[100000+5];
5 scanf("%d %d %d",&n,&l,&r);
6 for(int i=1;i<=n;i++) scanf("%d",&temp_a[i]);
7 for(int i=1;i<=n;i++) scanf("%d",&p[i]);
8 for(int i=1;i<=n;i++) a[ p[i] ]=temp_a[i];
9
10 //for(int i=1;i<=n;i++) printf("%d ",a[i]);printf("\n");
11
12 int now=l-a[1]-1;
13 for(int i=1;i<=n;i++)
14 {
15 if(l-a[i] > now) c[i]=(now=l-a[i]);
16 else c[i]=(now+=1);
17 if( c[i]+a[i] > r ){
18 printf("-1\n");
19 return 0;
20 }
21 }
22
23 for(int i=1;i<=n;i++){
24 if(i!=1) printf(" ");
25 printf("%d",c[p[i]]+a[p[i]]);
26 }
27 printf("\n");
28 return 0;
29 }
