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;
const int N = 100 * 1000 + 10, LG = 35;
long long mn[LG + 10][N], sum[LG + 10][N];
int par[LG + 10][N];
pair<long long, long long> get(int ind, long long k) {
  pair<long long, long long> ans(0,
                                 (1000LL * 1000 * 1000 * 1000 * 1000 * 1000));
  for (int i = 0; i <= LG; i++)
    if (k & (1LL << i)) {
      ans.first += sum[i][ind];
      ans.second = min(ans.second, mn[i][ind]);
      ind = par[i][ind];
    }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i <= n - 1; i++) cin >> par[0][i];
  for (int i = 0; i <= n - 1; i++) {
    cin >> sum[0][i];
    mn[0][i] = sum[0][i];
  }
  for (int i = 1; i <= LG; i++)
    for (int j = 0; j <= n - 1; j++) {
      par[i][j] = par[i - 1][par[i - 1][j]];
      mn[i][j] = min(mn[i - 1][j], mn[i - 1][par[i - 1][j]]);
      sum[i][j] = sum[i - 1][j] + sum[i - 1][par[i - 1][j]];
    }
  for (int i = 0; i <= n - 1; i++) {
    pair<long long, long long> ans = get(i, k);
    cout << ans.first << " " << ans.second << endl;
  }
  return 0;
}
