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;
int n, fa[100010][40], mn[100010][40];
long long k, s[100010][40];
void work(int x) {
  long long si = 0;
  int mi = 1e9;
  for (int i = 35; i >= 0; i--) {
    if ((1LL << i) & k) {
      si += s[x][i];
      mi = min(mi, mn[x][i]);
      x = fa[x][i];
    }
  }
  printf("%I64d %d\n", si, mi);
}
int main() {
  scanf("%d%I64d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &fa[i][0]);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &mn[i][0]);
    s[i][0] = mn[i][0];
  }
  for (int i = 1; i <= 35; i++) {
    for (int j = 0; j < n; j++) {
      fa[j][i] = fa[fa[j][i - 1]][i - 1];
      s[j][i] = s[j][i - 1] + s[fa[j][i - 1]][i - 1];
      mn[j][i] = min(mn[j][i - 1], mn[fa[j][i - 1]][i - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    work(i);
  }
  return 0;
}