Problem: You are given a functional graph. It is a directed graph, in which from each vertex goes exactly one arc. The vertices are numerated from 0 to n - 1.

Graph is given as the array f0, f1, ..., fn - 1, where fi — the number of vertex to which goes the only arc from the vertex i. Besides you are given array with weights of the arcs w0, w1, ..., wn - 1, where wi — the arc weight from i to fi.

<image> The graph from the first sample test.

Also you are given the integer k (the length of the path) and you need to find for each vertex two numbers si and mi, where:

  * si — the sum of the weights of all arcs of the path with length equals to k which starts from the vertex i; 
  * mi — the minimal weight from all arcs on the path with length k which starts from the vertex i. 



The length of the path is the number of arcs on this path.

Input

The first line contains two integers n, k (1 ≤ n ≤ 105, 1 ≤ k ≤ 1010). The second line contains the sequence f0, f1, ..., fn - 1 (0 ≤ fi < n) and the third — the sequence w0, w1, ..., wn - 1 (0 ≤ wi ≤ 108).

Output

Print n lines, the pair of integers si, mi in each line.

Examples

Input

7 3
1 2 3 4 3 2 6
6 3 1 4 2 2 3


Output

10 1
8 1
7 1
10 2
8 2
7 1
9 3


Input

4 4
0 1 2 3
0 1 2 3


Output

0 0
4 1
8 2
12 3


Input

5 3
1 2 3 4 0
4 1 2 14 3


Output

7 1
17 1
19 2
21 3
8 1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
const int LOGMAX = 36;
int N, log_K;
long long K;
int f[MAX], w[MAX];
int P[LOGMAX][MAX], M[LOGMAX][MAX];
long long S[LOGMAX][MAX];
long long sum_query(int k) {
  long long sum = 0;
  for (int i = int(log_K); i >= int(0); i--) {
    if (K & (1LL << i)) {
      sum += S[i][k];
      k = P[i][k];
    }
  }
  return sum;
}
int min_query(int k) {
  int minn = (int)1e9;
  for (int i = int(log_K); i >= int(0); i--) {
    if (K & (1LL << i)) {
      minn = min(minn, M[i][k]);
      k = P[i][k];
    }
  }
  return minn;
}
int main() {
  scanf("%d %lld", &N, &K);
  for (int i = int(0); i < int(N); i++) {
    scanf("%d", f + i);
  }
  for (int i = int(0); i < int(N); i++) {
    scanf("%d", w + i);
  }
  log_K = 63 - __builtin_clzll(K);
  memset(M, 0x7F, sizeof M);
  memset(P, -1, sizeof P);
  for (int i = int(0); i < int(N); i++) {
    P[0][i] = f[i];
    M[0][i] = w[i];
    S[0][i] = w[i];
  }
  for (int i = int(1); i < int(LOGMAX); i++) {
    for (int j = int(0); j < int(N); j++) {
      if (P[i - 1][j] != -1) {
        P[i][j] = P[i - 1][P[i - 1][j]];
        M[i][j] = min(M[i - 1][j], M[i - 1][P[i - 1][j]]);
        S[i][j] = S[i - 1][j] + S[i - 1][P[i - 1][j]];
      }
    }
  }
  for (int i = int(0); i < int(N); i++) {
    printf("%lld %d\n", sum_query(i), min_query(i));
  }
  return 0;
}