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;
inline int init() {
  int now = 0, ju = 1;
  char c;
  bool flag = false;
  while (1) {
    c = getchar();
    if (c == '-')
      ju = -1;
    else if (c >= '0' && c <= '9') {
      now = now * 10 + c - '0';
      flag = true;
    } else if (flag)
      return now * ju;
  }
}
inline long long llinit() {
  long long now = 0, ju = 1;
  char c;
  bool flag = false;
  while (1) {
    c = getchar();
    if (c == '-')
      ju = -1;
    else if (c >= '0' && c <= '9') {
      now = now * 10 + c - '0';
      flag = true;
    } else if (flag)
      return now * ju;
  }
}
long long k, sum[100005][42], mn[100005][42];
int n, f[100005][42];
void calc() {
  for (int j = 1; j <= 40; j++) {
    for (int i = 0; i < n; i++) {
      f[i][j] = f[f[i][j - 1]][j - 1];
      sum[i][j] = sum[f[i][j - 1]][j - 1] + sum[i][j - 1];
      mn[i][j] = min(mn[f[i][j - 1]][j - 1], mn[i][j - 1]);
    }
  }
}
pair<long long, long long> query(int now) {
  long long ret = 0x7f7f7f7f;
  long long res = 0;
  for (int i = 40; i >= 0; i--) {
    if ((1LL << i) & k) {
      ret = min(ret, mn[now][i]);
      res = res + sum[now][i];
      now = f[now][i];
    }
  }
  return make_pair(res, ret);
}
int main() {
  n = init();
  k = llinit();
  for (int i = 0; i < n; i++) {
    f[i][0] = init();
  }
  for (int i = 0; i < n; i++) {
    sum[i][0] = mn[i][0] = llinit();
  }
  calc();
  pair<long long, long long> tmp;
  for (int i = 0; i < n; i++) {
    tmp = query(i);
    printf("%I64d %I64d\n", tmp.first, tmp.second);
  }
  return 0;
}