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 = 1e5 + 5, L = 40;
long long s[N][L], p[N][L], m[N][L];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> p[i][0];
  }
  for (int i = 0; i < n; i++) {
    cin >> s[i][0];
    m[i][0] = s[i][0];
  }
  for (int j = 1; j < L; j++) {
    for (int i = 0; i < n; i++) {
      p[i][j] = p[p[i][j - 1]][j - 1];
      s[i][j] = s[i][j - 1] + s[p[i][j - 1]][j - 1];
      m[i][j] = min(m[i][j - 1], m[p[i][j - 1]][j - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    long long sum = 0, mn = 1e18, u = i, x = k, c = 0;
    while (x > 0) {
      if (x & 1) {
        sum += s[u][c];
        mn = min(mn, m[u][c]);
        u = p[u][c];
      }
      x >>= 1;
      c++;
    }
    cout << sum << " " << mn << "\n";
  }
}
