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;
long long fa[51][200007], sum[51][200007], mn[51][200007];
inline long long read() {
  long long x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * flag;
}
signed main() {
  long long n = read(), k = read();
  for (long long i = 0; i <= n - 1; ++i) fa[0][i] = read();
  for (long long i = 0; i <= n - 1; ++i) {
    long long x = read();
    sum[0][i] = x;
    mn[0][i] = x;
  }
  for (long long i = 1; i <= 50; ++i) {
    for (long long j = 0; j <= n - 1; ++j) {
      fa[i][j] = fa[i - 1][fa[i - 1][j]];
      sum[i][j] = sum[i - 1][j] + sum[i - 1][fa[i - 1][j]];
      mn[i][j] = min(mn[i - 1][j], mn[i - 1][fa[i - 1][j]]);
    }
  }
  for (long long i = 0; i <= n - 1; ++i) {
    long long res = 0, minn = mn[0][i], p = i;
    for (long long j = 0; j <= 51; ++j) {
      if (k & (1LL << j)) {
        res += sum[j][p];
        minn = min(minn, mn[j][p]);
        p = fa[j][p];
      }
    }
    printf("%lld %lld\n", res, minn);
  }
  return 0;
}