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, m, to[100005], mi[100005][35], v[100005], ppow[40];
long long ans, bz[100005][35], s[100005][35], sum, minn, step;
inline long long read() {
  long long ret = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    ret = (ret << 3) + (ret << 1) + (ch ^ 48);
    ch = getchar();
  }
  return ret * ff;
}
void solve(long long x) {
  minn = (1e18), sum = 0, step = 0;
  for (int i = 34; i >= 0; i--) {
    if (step + ppow[i] <= m) {
      sum += s[x][i];
      minn = min(minn, mi[x][i]);
      step += ppow[i];
      x = bz[x][i];
    }
  }
}
int main() {
  ppow[0] = 1;
  for (int i = 1; i <= 34; i++) ppow[i] = ppow[i - 1] * 2;
  n = read(), m = read();
  for (int i = 0; i < n; i++) to[i] = read(), bz[i][0] = to[i];
  for (int i = 0; i < n; i++) v[i] = read(), mi[i][0] = v[i], s[i][0] = v[i];
  for (int k = 1; k <= 34; k++)
    for (int i = 0; i < n; i++) {
      bz[i][k] = bz[bz[i][k - 1]][k - 1];
      mi[i][k] = min(mi[i][k - 1], mi[bz[i][k - 1]][k - 1]);
      s[i][k] = s[i][k - 1] + s[bz[i][k - 1]][k - 1];
    }
  for (int i = 0; i < n; i++) {
    solve(i);
    printf("%lld %lld\n", sum, minn);
  }
  return 0;
}
