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 st {
  long long cur;
  long long len;
  int mini;
  int dest;
};
int n;
long long k;
vector<st> to[100100];
st res[100100];
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> res[i].dest;
  for (int i = 0; i < n; i++) cin >> res[i].len;
  for (int i = 0; i < n; i++) {
    res[i].mini = res[i].len;
    res[i].cur = 1LL;
    to[i].push_back(res[i]);
  }
  while (res[0].cur < k) {
    int s = to[0].size() - 1;
    if (res[0].cur + to[0][s].cur > k) {
      for (int i = 0; i < n; i++) to[i].pop_back();
      continue;
    }
    for (int i = 0; i < n; i++) {
      res[i].cur += to[i][s].cur;
      res[i].len += to[res[i].dest][s].len;
      res[i].mini = min(res[i].mini, to[res[i].dest][s].mini);
      res[i].dest = to[res[i].dest][s].dest;
    }
    if (res[0].cur == 2 * to[0][s].cur) {
      for (int i = 0; i < n; i++) to[i].push_back(res[i]);
    }
  }
  for (int i = 0; i < n; i++) cout << res[i].len << " " << res[i].mini << "\n";
  return 0;
}
