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;
struct node {
  long long int next, sum, mn;
};
vector<node> operate(vector<node> a, vector<node> b) {
  vector<node> res;
  res.resize(b.size());
  if (a.size() == 0) return b;
  for (long long int i = 0; i < a.size(); i++) {
    res[i].next = a[b[i].next].next;
    res[i].sum = a[b[i].next].sum + b[i].sum;
    res[i].mn = min(a[b[i].next].mn, b[i].mn);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, k;
  cin >> n >> k;
  vector<node> v;
  v.resize(n);
  for (long long int i = 0; i < n; i++) cin >> v[i].next;
  for (long long int i = 0; i < n; i++) cin >> v[i].sum, v[i].mn = v[i].sum;
  vector<node> ans;
  while (k) {
    if (k & 1) ans = operate(ans, v);
    v = operate(v, v);
    k >>= 1;
  }
  for (auto i : ans) cout << i.sum << " " << i.mn << '\n';
}