Problem: 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.

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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].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline void scf(int &x) { scanf("%d", &x); }
inline void scf(long long &x) { scanf("%I64d", &x); }
inline void scf(double &x) { scanf("%lf", &x); }
inline void scf(char &x) { scanf("%c", &x); }
inline void scf(char *x) { scanf("%s", x); }
const int N = 1e5 + 7;
const int mars = 0x3c3c3c3c;
typedef int Arr[N];
typedef char Err[N];
int m, n, u, v, w, a, b, l, t, r, q, s, x, y, cnt;
Arr A, B, C, T;
inline bool Comp(const int &a, const int &b) {
  if (C[a] == C[b])
    return A[a] < A[b];
  else
    return C[a] > C[b];
}
int main() {
  int i, j, k;
  scf(n), scf(l), scf(r);
  for (i = 1; i <= n; i++) scf(A[i]);
  for (i = 1; i <= n; i++) scf(C[i]);
  for (i = 1; i <= n; i++) T[i] = i;
  sort(T + 1, T + n + 1, Comp);
  x = r - A[T[1]], B[T[1]] = A[T[1]] + x;
  for (i = 2; i <= n; i++) {
    j = T[i];
    if (C[T[i - 1]] != C[j]) x--;
    if (A[j] + x < l) {
      printf("-1");
      return 0;
    }
    if (A[j] + x > r)
      B[j] = r, x = r - A[j];
    else
      B[j] = A[j] + x;
  }
  for (i = 1; i <= n; i++) printf("%d ", B[i]);
}