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 = 123456789987654321;
long long int N, K;
vector<long long int> F, W;
vector<vector<long long int> > par, minw, sumw;
int main() {
  scanf("%I64d %I64d", &N, &K);
  F = W = vector<long long int>(N);
  for (int i = 0; i < N; i++) {
    scanf("%I64d", &F[i]);
  }
  for (int i = 0; i < N; i++) {
    scanf("%I64d", &W[i]);
  }
  par = minw = sumw =
      vector<vector<long long int> >(35, vector<long long int>(N, -1));
  for (int i = 0; i < 35; i++) {
    for (int j = 0; j < N; j++) {
      if (i == 0) {
        par[i][j] = F[j];
        minw[i][j] = W[j];
        sumw[i][j] = W[j];
      } else {
        if (par[i - 1][j] == -1) continue;
        par[i][j] = par[i - 1][par[i - 1][j]];
        minw[i][j] = min(minw[i - 1][j], minw[i - 1][par[i - 1][j]]);
        sumw[i][j] = sumw[i - 1][j] + sumw[i - 1][par[i - 1][j]];
      }
    }
  }
  for (int i = 0; i < N; i++) {
    int now = i;
    long long int sum = 0, mini = inf;
    for (int j = 35; j--;) {
      if ((((long long int)1 << j) & K)) {
        sum += sumw[j][now];
        mini = min(mini, minw[j][now]);
        now = par[j][now];
      }
    }
    printf("%I64d %I64d\n", sum, mini);
  }
}