time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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].

【题目链接】:

【题解】

因为

c[i] = b[i]-a[i];

题目所给的p[i]实际上是c[i]之间的大小关系;

p[i]=1是最小的,p[i]=2则是第二小…p[i]=n则是最大的;

这样我们可以让

c的值从小到大为

c[1],c[1]+1,c[1]+2,c[1]+3…c[1]+n-1;

即每个值递增1

这样能够在满足p[]数组的情况下尽可能地让c[i]小,相应的b[i]也就小了

但是有可能我们要求c[i]为递增的规律的时候;

b[i]是根据c[i]求出来的

某个i会出现

b[i]< l

这个时候让b[i]为l(b数组没有说一定要全都不同)

然后c[i]相应地调整;

c[i]会变大一点,这样能保证满足p数组,同时b[i]还是最小的l,所以还是能够保证c[i]此时也是最小的

(但是如果b[i]>r,则直接输出无解;

因为此前的操作我们已经尽可能的让c数组小了;

如果还不行(即b[i]还是太大),那么就无解了;)

然后再从i+1开始

让

c[i+1]=c[i]+1

c[i+2] = c[i]+2

…

即此后每个值还是递增1

然后求出相应的b数组

(所以这个c数组最后可能中间会有不是c[i]=c[i-1]+1的断层);

一开始的时候设最小的c的下标为idx

让b[idx]=l;(这样能保证b数组、c数组一开始是最小的了)

求出这个时候的c[idx]作为发生器;

然后做上面的过程就好;

贪心点就是时刻保证c、b数组是最小的,这样b数组在某个时刻通过c数组获得的时候,如果大于r则肯定无解;

小于l则可以调整为l,还是保证c、b为最小的；

【完整代码】

#include 
     
      using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long long#define rep1(i,a,b) for (int i = a;i <= b;i++)#define rep2(i,a,b) for (int i = a;i >= b;i--)#define mp make_pair#define pb push_back#define fi first#define se second#define rei(x) scanf("%d",&x)#define rel(x) scanf("%I64d",&x)typedef pair
      
        pii;typedef pair
       
         pll;const int dx[9] = {
    0,1,-1,0,0,-1,-1,1,1};const int dy[9] = {
    0,0,0,-1,1,-1,1,-1,1};const double pi = acos(-1.0);const int MAXN = 1e5+100;int a[MAXN],n,l,r,cc[MAXN],b[MAXN],c[MAXN];void wujie(){    puts("-1");    exit(0);}int main(){    //freopen("F:\\rush.txt","r",stdin);    cin >> n >> l >> r;    rep1(i,1,n)        rei(a[i]);    rep1(i,1,n)    {        int x;        rei(x);        cc[x] = i;    }    //c[i] = b[i]-a[i];    //c[idx]>c[1]    //b[idx]-a[idx]>b[1]-a[1];    //c[i]+a[i] = b[i];    b[cc[1]] = l;    c[cc[1]] = b[cc[1]]-a[cc[1]];    int temp = b[cc[1]]-a[cc[1]];    rep1(i,2,n)    {        int idx = cc[i];        //        c[idx] = temp+1;        b[idx] = c[idx]+a[idx];        temp++;        if (b[idx]
        
         r)            wujie();    }    for (int i = 1;i <= n;i++)    {        printf("%d",b[i]);        if (i==n)            puts("");        else            putchar(' ');    }    return 0;}