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, m, p[200010][40];
long long k, ans1, ans2, sum[200010][40], Min[200010][40];
void init() {
  for (int i = 1; i < 35; i++)
    for (int j = 0; j < n; j++)
      p[j][i] = p[p[j][i - 1]][i - 1],
      Min[j][i] = min(Min[j][i - 1], Min[p[j][i - 1]][i - 1]),
      sum[j][i] = sum[j][i - 1] + sum[p[j][i - 1]][i - 1];
}
void query(int x, long long y) {
  ans1 = 0;
  ans2 = (long long)10e10;
  long long k;
  int z = x;
  for (int i = 34; i >= 0; i--) {
    k = 1LL << i;
    if (y >= k) {
      y -= k;
      ans1 += sum[z][i];
      ans2 = min(Min[z][i], ans2);
      z = p[z][i];
    }
  }
}
int main() {
  cin >> n >> k;
  int x;
  for (int i = 0; i < n; i++) scanf("%d", &p[i][0]);
  for (int i = 0; i < n; i++) scanf("%I64d", &sum[i][0]), Min[i][0] = sum[i][0];
  init();
  for (int i = 0; i < n; i++) query(i, k), printf("%I64d %I64d\n", ans1, ans2);
  return 0;
}