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;
const double eps = 1e-15;
const double pi = acos(-1.0);
const int inf = 1 << 30;
const long long Mod = 1000000007;
const int N = 1e5 + 5;
const int Log = 50;
long long Sum[N][Log];
long long Min[N][Log];
long long Par[N][Log];
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> Par[i][0];
  }
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    Sum[i][0] = Min[i][0] = x;
  }
  for (int j = 1; j < Log; j++) {
    for (int i = 0; i < n; i++) {
      Par[i][j] = Par[Par[i][j - 1]][j - 1];
      Sum[i][j] = Sum[i][j - 1] + Sum[Par[i][j - 1]][j - 1];
      Min[i][j] = min(Min[i][j - 1], Min[Par[i][j - 1]][j - 1]);
    }
  }
  for (long long i = 0; i < n; i++) {
    long long s = 0;
    long long Mn = Mod;
    long long cur = i;
    for (int j = 0; j < Log; j++) {
      if (k & (1LL << j)) {
        Mn = min(Mn, Min[cur][j]);
        s += Sum[cur][j];
        cur = Par[cur][j];
      }
    }
    cout << s << ' ' << Mn << '\n';
  }
  return 0;
}