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;
long long n, k, f[100005][35], m[100005][35], s[100005][35];
void build(bool x) {
  for (int b = 1; b < 35; b++) {
    for (int i = 0; i < n; i++) {
      if (x)
        m[i][b] = min(m[i][b - 1], m[f[i][b - 1]][b - 1]);
      else
        s[i][b] = s[i][b - 1] + s[f[i][b - 1]][b - 1];
    }
  }
}
long long query(long long l, long long len, bool x) {
  long long a1 = 1e10, a2 = 0;
  for (int b = 34; b >= 0; b--) {
    long long z = (1LL) << b;
    if (len >= z) {
      if (x)
        a1 = min(a1, m[l][b]);
      else
        a2 += s[l][b];
      l = f[l][b];
      len -= z;
    }
  }
  return ((x) ? a1 : a2);
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> f[i][0];
  for (int i = 0; i < n; i++) cin >> m[i][0], s[i][0] = m[i][0];
  for (int b = 1; b < 35; b++)
    for (int i = 0; i < n; i++) m[i][b] = 1e9;
  for (int b = 1; b < 35; b++)
    for (int i = 0; i < n; i++) f[i][b] = f[f[i][b - 1]][b - 1];
  build(0), build(1);
  for (int i = 0; i < n; i++)
    cout << query(i, k, 0) << " " << query(i, k, 1) << endl;
}
