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 maxn = 1e5 + 10;
long long dp1[35][maxn], dp2[35][maxn], fa[35][maxn];
long long n, k;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> fa[0][i];
  for (int i = 0; i < n; i++) {
    cin >> dp1[0][i];
    dp2[0][i] = dp1[0][i];
  }
  for (int i = 1; (1LL << i) <= k; i++) {
    for (int j = 0; j < n; j++) {
      dp1[i][j] = dp1[i - 1][j] + dp1[i - 1][fa[i - 1][j]];
      dp2[i][j] = min(dp2[i - 1][j], dp2[i - 1][fa[i - 1][j]]);
      fa[i][j] = fa[i - 1][fa[i - 1][j]];
    }
  }
  for (int i = 0; i < n; i++) {
    int u = i;
    long long sum = 0, Min = 0x7f7f7f7f7f;
    for (int j = 0; j < 35; j++) {
      if (k >> j & 1) {
        sum += dp1[j][u];
        Min = min(Min, dp2[j][u]);
        u = fa[j][u];
      }
    }
    cout << sum << " " << Min << endl;
  }
}
