D. Dasha and Very Difficult Problem----思维题

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 sequencec = [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 51 1 1 1 13 1 5 4 2

output

3 1 5 4 2 

input

4 2 93 4 8 93 2 1 4

output

2 2 2 9 

input

6 1 51 1 1 1 1 12 3 5 4 1 6

output

-1

Note

Sequence b which was found in the second sample is suitable, because calculated sequencec = [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].

题目链接：http://codeforces.com/contest/761/problem/D

题目的意思是说有个a[i]数组，和一个b[i]数组，我们定义c[i]=b[i]-a[i]，p[i]数组是c数组的相对大小，现给你a和p，让你把b数组求出来。

我刚开始敲了一个我以为超时的代码，但是并没有超时，wa到了第15组，我看了第15组的数据，还是没有看出哪里错了，大牛们路过的时候帮我看看哪里错了吧。

我看了my room 第一的代码，好像这样写确实有道理，我们已知c[i]=b[i]-a[i],所以b[i]=a[i]+c[i]，而p数组是c的相对大小，我本想找组数据推翻这样写，但是我并没有找出数据，这样写是对的，比较尴尬。。。

代码：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int a[200000];int p[200000];long long b[200000];int main(){    int n,l,r;    scanf("%d%d%d",&n,&l,&r);    for(int i=0;i<n;i++){        scanf("%d",&a[i]);    }    for(int i=0;i<n;i++){        scanf("%d",&p[i]);    }    long long mix=l;    for(int i=0;i<n;i++){        b[i]=a[i]+p[i];        mix=max(mix,b[i]);    }    if(mix>r){        int h=mix-=r;        for(int i=0;i<n;i++){            b[i]-=h;        }    }    bool flag=false;    for(int i=0;i<n;i++){        if(b[i]<l){            flag=true;            break;        }    }    if(flag){        cout<<"-1"<<endl;    }    else{        for(int i=0;i<n;i++){            if(i!=0)                cout<<" "<<b[i];            else                cout<<b[i];        }        cout<<endl;    }    return 0;}

我一开始wa到第15组的代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;struct node{    int a,p,k,b;}xin[200000];bool cmp(struct node a,struct node b){    return a.p>b.p;}bool cmp2(struct node a,struct node b){    return a.k<b.k;}int c[200000];int main(){    int n,l,r;    scanf("%d%d%d",&n,&l,&r);    for(int i=0;i<n;i++){        scanf("%d",&xin[i].a);        xin[i].k=i;    }    for(int i=0;i<n;i++){        scanf("%d",&xin[i].p);    }    int k=r;    sort(xin,xin+n,cmp);    xin[0].b=k;    c[0]=k-xin[0].a;    bool flag=false;    for(int i=1;i<n;i++){        while(k>=l){            if(k-xin[i].a<c[i-1]){                xin[i].b=k;                c[i]=k-xin[i].a;                break;            }            k--;//感觉这里TLE了，但是竟然wa了        }        if(k<l){            flag=true;            break;        }    }    if(flag){        cout<<"-1"<<endl;    }    else{        sort(xin,xin+n,cmp2);        for(int i=0;i<n;i++){            printf(i==0?"%d":" %d",xin[i].b);        }        cout<<endl;    }    return 0;}