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;
long long n, k, ans[100001][51];
long long Min[100001][51], son[100001][51], er[41], num, l[41];
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> son[i][0];
  for (int i = 0; i < n; i++) {
    cin >> ans[i][0];
    Min[i][0] = ans[i][0];
  }
  for (int i = 1; i <= 40; i++) {
    for (int j = 0; j < n; j++) {
      son[j][i] = son[son[j][i - 1]][i - 1];
    }
  }
  for (int i = 1; i <= 40; i++)
    for (int j = 0; j < n; j++) {
      Min[j][i] = min(Min[j][i - 1], Min[son[j][i - 1]][i - 1]);
      ans[j][i] = ans[j][i - 1] + ans[son[j][i - 1]][i - 1];
    }
  er[0] = 1;
  for (int i = 1; i <= 40; i++) er[i] = er[i - 1] * 2;
  for (int i = 40; i >= 0; i--) {
    if (k >= er[i]) {
      num++;
      l[num] = i;
      k = k - er[i];
    }
  }
  for (int i = 0; i < n; i++) {
    long long x = i;
    long long minn = 1000000000000;
    long long anss = 0;
    for (int j = 1; j <= num; j++) {
      minn = min(minn, Min[x][l[j]]);
      anss = anss + ans[x][l[j]];
      x = son[x][l[j]];
    }
    cout << anss << "  " << minn << endl;
  }
}