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;
long long k;
int mi[100007];
int dp[40][100007];
long long sum[40][100007];
int mini[40][100007];
long long si[100007];
void func() {
  for (int i = 1; i < 40; i++)
    for (int j = 0; j < n; j++) dp[i][j] = dp[i - 1][dp[i - 1][j]];
  for (int i = 1; i < 40; i++)
    for (int j = 0; j < n; j++) {
      sum[i][j] = sum[i - 1][j] + sum[i - 1][dp[i - 1][j]];
      mini[i][j] = min(mini[i - 1][j], mini[i - 1][dp[i - 1][j]]);
    }
  for (int i = 0; i < n; i++) {
    mi[i] = INT_MAX;
    long long temp = k;
    int j = i;
    while (temp) {
      long long t = log2(temp);
      si[i] += sum[t][j];
      mi[i] = min(mi[i], mini[t][j]);
      j = dp[t][j];
      temp -= (1LL << t);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0, f; i < n; i++) {
    cin >> f;
    dp[0][i] = f;
  }
  for (int i = 0, w; i < n; i++) {
    cin >> w;
    sum[0][i] = w;
    mini[0][i] = w;
  }
  func();
  for (int i = 0; i < n; i++) cout << si[i] << " " << mi[i] << endl;
  return 0;
}