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【Codeforces 761 D Dasha and Very Difficult Problem】

2019-11-14 11:00:59
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D. Dasha and Very Difficult PRoblem 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

Note

Sequence b which was found in the second sample is suitable, because calculated sequence c?=?[2?-?3,?2?-?4,?2?-?8,?9?-?9]?=?[?-?1,??-?2,??-?6,?0] (note that ci?=?bi?-?ai) has compressed sequence equals to p?=?[3,?2,?1,?4].

依據(jù)Ci的排序sort,令B1 = L,C1 = L - A1,Ci = C i-1 + 1 , Bi = Ai + Ci,若Bi < L,可令Bi = L,Ci = L - Ai,若Bi > R ,則無解,反之,記錄一開始的位置,還原輸出Bi;

AC代碼:

#include<cstdio>#include<algorithm>using namespace std;struct node{ int a,b,c,p,n;}st[100010];bool cmp(node i,node j){ return i.p < j.p; }bool cnp(node i,node j){ return i.n < j.n; }int main(){ int N,L,R; scanf("%d %d %d",&N,&L,&R); for(int i = 1 ; i <= N; i++) scanf("%d",&st[i].a),st[i].n = i; for(int i = 1 ; i <= N; i++) scanf("%d",&st[i].p); sort(st + 1, st + 1 + N,cmp); st[1].b = L,st[1].c = L - st[1].a; for(int i = 2 ; i <= N; i++){ st[i].c = st[i - 1].c + 1; st[i].b = st[i].c + st[i].a; if(st[i].b < L) st[i].b = L,st[i].c = L - st[i].a; if(st[i].b > R){ printf("-1/n"); return 0; } } sort(st + 1, st + 1 + N,cnp); for(int i = 1 ; i <= N; i++) printf("%d ",st[i].b); return 0;}
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