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;
int n;
long long k, sum[100009][49], mn[100009][49];
int nxt[100009][49];
pair<long long, long long> ch(long long x) {
  pair<long long, long long> ret;
  ret.first = 1e9;
  for (long long j = 1, i = 0; j <= k; j *= 2, i++) {
    if (j & k) {
      ret.first = min(ret.first, mn[x][i]);
      ret.second += sum[x][i];
      x = nxt[x][i];
    }
  }
  return ret;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> nxt[i][0];
  for (int i = 0; i < n; i++) {
    cin >> sum[i][0];
    mn[i][0] = sum[i][0];
  }
  for (int j = 1; j < 34; j++) {
    for (int i = 0; i < n; i++) {
      nxt[i][j] = nxt[nxt[i][j - 1]][j - 1];
      sum[i][j] = sum[i][j - 1] + sum[nxt[i][j - 1]][j - 1];
      mn[i][j] = min(mn[i][j - 1], mn[nxt[i][j - 1]][j - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    pair<long long, long long> p = ch(i);
    cout << p.second << " " << p.first << "\n";
  }
}
