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
#include <bits/stdc++.h>
using namespace std;
struct node {
  int to, min = INT_MAX;
  long long sum = 0;
};
vector<int> t(int(1e5 + 9)), w(int(1e5 + 9));
vector<vector<node>> bs(40, vector<node>(int(1e5 + 9)));
void solve(int x, long long k) {
  int min_ = INT_MAX;
  long long sum = 0;
  for (int i = 0; i < 35; i++) {
    if (!(k & (1LL << i))) continue;
    sum += bs[i][x].sum;
    min_ = min(min_, bs[i][x].min);
    x = bs[i][x].to;
  }
  cout << sum << " " << min_ << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> t[i];
  for (int i = 0; i < n; i++) cin >> w[i];
  for (int i = 0; i < n; i++) {
    bs[0][i].to = t[i];
    bs[0][i].sum = w[i];
    bs[0][i].min = w[i];
  }
  for (int i = 1; i < 40; i++)
    for (int j = 0; j < n; j++) {
      bs[i][j].to = bs[i - 1][bs[i - 1][j].to].to;
      bs[i][j].sum = bs[i - 1][j].sum + bs[i - 1][bs[i - 1][j].to].sum;
      bs[i][j].min = min(bs[i - 1][j].min, bs[i - 1][bs[i - 1][j].to].min);
    }
  for (int i = 0; i < n; i++) solve(i, k);
  return 0;
}
