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;
const int MX = 35;
int n;
long long len, ans1, ans2, k, pred[50][100000 + 100], mn[50][100000 + 100],
    suma[50][100000 + 100];
void init_() {
  for (int i = 1; i <= MX; i++)
    for (int j = 0; j < n; j++) {
      pred[i][j] = pred[i - 1][pred[i - 1][j]];
      suma[i][j] = suma[i - 1][pred[i - 1][j]] + suma[i - 1][j];
      mn[i][j] = min(mn[i - 1][j], mn[i - 1][pred[i - 1][j]]);
    }
}
void solve(int x, long long y) {
  for (int i = len; i >= 0; i--) {
    if (1LL << i <= y) {
      ans1 += suma[i][x];
      ans2 = min(ans2, mn[i][x]);
      x = pred[i][x];
      y -= 1LL << i;
    }
  }
}
int main() {
  cin >> n >> k;
  len = log2(k);
  for (int i = 1; i <= n; i++) scanf("%lld", &pred[0][i - 1]);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &suma[0][i - 1]);
    mn[0][i - 1] = suma[0][i - 1];
  }
  init_();
  for (int i = 0; i < n; i++) {
    ans1 = 0;
    ans2 = 1e12;
    solve(i, k);
    printf("%lld %lld\n", ans1, ans2);
  }
  return 0;
}