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;
int dp[101000][40], Min[101000][40];
long long sum[101000][40], pow2[40];
int main() {
  pow2[0] = 1;
  for (int i = 1; i < 40; i++) pow2[i] = pow2[i - 1] << 1;
  int n, x;
  long long k;
  cin >> n >> k;
  memset(dp, 0, sizeof(dp));
  memset(Min, 0x3f3f3f3f, sizeof(Min));
  memset(sum, 0, sizeof(sum));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    x++;
    dp[i][0] = x;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    Min[i][0] = sum[i][0] = x;
  }
  for (int i = 1; i < 40; i++)
    for (int j = 1; j <= n; j++) {
      dp[j][i] = dp[dp[j][i - 1]][i - 1];
      sum[j][i] = sum[j][i - 1] + sum[dp[j][i - 1]][i - 1];
      Min[j][i] = min(Min[j][i - 1], Min[dp[j][i - 1]][i - 1]);
    }
  for (int i = 1; i <= n; i++) {
    long long ans = 0, mint = 0x3f3f3f3f;
    for (int p = 40, x = i; p >= 0; p--)
      if (pow2[p] & k) {
        ans += sum[x][p];
        mint = min(mint, (long long)Min[x][p]);
        x = dp[x][p];
      }
    printf("%I64d %I64d\n", ans, mint);
  }
  return 0;
}