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 N = 100001;
const int Log = 40;
int f[N][Log];
long long sumw[N][Log];
int minw[N][Log];
int main() {
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> f[i][0];
  for (int i = 0; i < n; i++) {
    cin >> sumw[i][0];
    minw[i][0] = sumw[i][0];
  }
  for (int i = 1; i < Log; i++)
    for (int j = 0; j < n; j++) {
      f[j][i] = f[f[j][i - 1]][i - 1];
      sumw[j][i] = sumw[j][i - 1] + sumw[f[j][i - 1]][i - 1];
      minw[j][i] = min(minw[j][i - 1], minw[f[j][i - 1]][i - 1]);
    }
  for (int i = 0; i < n; i++) {
    long long res_sumw = 0;
    int res_minw = 1e9 + 1;
    int v = i;
    for (int j = 0; j < Log; j++)
      if ((k >> j) & 1) {
        res_sumw += sumw[v][j];
        res_minw = min(res_minw, minw[v][j]);
        v = f[v][j];
      }
    cout << res_sumw << " " << res_minw << endl;
  }
  return 0;
}
