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 long long int INF = 10000000000000;
const int MAXN = 100010;
const int MAXLOG = 50;
long long int par[MAXN][MAXLOG];
long long int ans[MAXN][MAXLOG];
long long int mini[MAXN][MAXLOG];
pair<long long int, long long int> solve(long long int a, long long int b) {
  long long int ret1 = 0, ret2 = INF;
  long long int now = a;
  vector<long long int> bini;
  while (b != 0) {
    bini.push_back(b % 2);
    b /= 2;
  }
  while (bini.size() != MAXLOG) bini.push_back(0);
  for (int j = MAXLOG - 1; j >= 0; j--) {
    if (bini[j] == 1) {
      ret1 += ans[now][j];
      ret2 = min(ret2, mini[now][j]);
      now = par[now][j];
    }
  }
  return make_pair(ret1, ret2);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ios::sync_with_stdio(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  for (int j = 0; j < MAXLOG; j++) {
    for (int i = 0; i < n; i++) {
      mini[i][j] = INF;
    }
  }
  for (int i = 0; i < n; i++) {
    cin >> par[i][0];
  }
  for (int i = 0; i < n; i++) {
    cin >> ans[i][0];
    mini[i][0] = ans[i][0];
  }
  for (int j = 1; j < MAXLOG; j++) {
    for (int i = 0; i < n; i++) {
      par[i][j] = par[par[i][j - 1]][j - 1];
    }
  }
  for (int j = 1; j < MAXLOG; j++) {
    for (int i = 0; i < n; i++) {
      ans[i][j] = ans[i][j - 1] + ans[par[i][j - 1]][j - 1];
    }
  }
  for (int j = 1; j < MAXLOG; j++) {
    for (int i = 0; i < n; i++) {
      mini[i][j] = min(mini[i][j - 1], mini[par[i][j - 1]][j - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    pair<long long int, long long int> answer = solve(i, k);
    cout << answer.first << ' ' << answer.second << endl;
  }
}