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;
int par[40][100010];
long long cst[100010];
long long mn[40][100010];
long long sum[40][100010];
int main() {
  int n;
  long long m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> par[0][i];
  for (int i = 0; i < n; i++) cin >> cst[i];
  for (int i = 0; i < n; i++) sum[0][i] = mn[0][i] = cst[i];
  for (int i = 1; i < 40; i++) {
    for (int j = 0; j < n; j++) {
      par[i][j] = par[i - 1][par[i - 1][j]];
      mn[i][j] = min(mn[i - 1][j], mn[i - 1][par[i - 1][j]]);
      sum[i][j] = sum[i - 1][j] + sum[i - 1][par[i - 1][j]];
    }
  }
  for (int i = 0; i < n; i++) {
    long long t = m, now = i, r = 0;
    long long ans1 = 100000000, ans2 = 0;
    while (t > 0) {
      if (t % 2) {
        ans2 += sum[r][now];
        ans1 = min(ans1, mn[r][now]);
        now = par[r][now];
      }
      r++;
      t /= 2;
    }
    cout << ans2 << " " << ans1 << endl;
  }
  return 0;
}
