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 long long INF = 1e15 + 6;
const int MAX_N = 1e5 + 6;
const long long MAX_K = 1e10 + 5;
const int MAX_P = 40;
struct Node {
  int to;
  long long len;
  long long min;
} node[MAX_P][MAX_N];
int binary[MAX_P];
Node MP(int _to, long long _len, long long _min) {
  Node tmp;
  tmp.to = _to;
  tmp.len = _len;
  tmp.min = _min;
  return tmp;
}
pair<long long, long long> query(int id, long long k) {
  memset(binary, 0, sizeof(binary));
  int i = 0;
  while (k > 0) {
    binary[i++] = k % 2;
    k /= 2;
  }
  pair<long long, long long> ret = make_pair(0, INF);
  for (int x = MAX_P - 1; x >= 0; x--) {
    if (binary[x] == 1) {
      ret = make_pair(ret.first + node[x][id].len,
                      min(ret.second, node[x][id].min));
      id = node[x][id].to;
    }
  }
  return ret;
}
int main() {
  int n;
  long long k;
  while (scanf("%d %I64d", &n, &k) != EOF) {
    for (int x = 0; n > x; x++) {
      scanf("%d", &node[0][x].to);
    }
    for (int x = 0; n > x; x++) {
      scanf("%d", &node[0][x].len);
      node[0][x].min = node[0][x].len;
    }
    for (int i = 1; MAX_P > i; i++) {
      for (int j = 0; n > j; j++) {
        int tmp = node[i - 1][j].to;
        node[i][j] =
            MP(node[i - 1][tmp].to, node[i - 1][j].len + node[i - 1][tmp].len,
               min(node[i - 1][j].min, node[i - 1][tmp].min));
      }
    }
    for (int x = 0; n > x; x++) {
      pair<long long, long long> ret = query(x, k);
      printf("%I64d %I64d\n", ret.first, ret.second);
    }
  }
}
