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 int kN = 100 * 1000;
const int kLog = 35;
int f[kN];
long long w[kN];
int up[kN][kLog];
long long mn[kN][kLog];
long long sum[kN][kLog];
int main() {
  std::ios_base::sync_with_stdio(false);
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> f[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> w[i];
  }
  for (int i = 0; i < n; ++i) {
    up[i][0] = f[i];
    mn[i][0] = sum[i][0] = w[i];
  }
  for (int h = 1; h < kLog; ++h) {
    for (int i = 0; i < n; ++i) {
      mn[i][h] = min(mn[i][h - 1], mn[up[i][h - 1]][h - 1]);
      sum[i][h] = sum[i][h - 1] + sum[up[i][h - 1]][h - 1];
      up[i][h] = up[up[i][h - 1]][h - 1];
    }
  }
  for (int i = 0; i < n; ++i) {
    long long sum_ = 0;
    long long mn_ = 1e9;
    int u = i;
    for (int b = 0; b < kLog; ++b) {
      if (k & (1LL << b)) {
        sum_ += sum[u][b];
        mn_ = min(mn_, mn[u][b]);
        u = up[u][b];
      }
    }
    cout << sum_ << ' ' << mn_ << '\n';
  }
  return 0;
}
