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;
signed main() {
  long long n, k;
  cin >> n >> k;
  long long a[n], b[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) cin >> b[i];
  pair<pair<long long, long long>, long long> dp[n + 1][40];
  for (long long i = 0; i < n; i++) {
    dp[i][0] = {{b[i], b[i]}, a[i]};
  }
  for (long long j = 1; j <= 38; j++) {
    for (long long i = 0; i < n; i++) {
      pair<pair<long long, long long>, long long> pp =
          dp[dp[i][j - 1].second][j - 1];
      dp[i][j] = {{dp[i][j - 1].first.first + pp.first.first,
                   min(pp.first.second, dp[i][j - 1].first.second)},
                  pp.second};
    }
  }
  for (long long i = 0; i < n; i++) {
    long long sum = 0, minn = 1e12, c = 0, curr = i;
    for (long long j = 38; j >= 0; j--) {
      if (c + (1LL << j) <= k) {
        c += (1LL << j);
        sum += dp[curr][j].first.first;
        minn = min(dp[curr][j].first.second, minn);
        curr = dp[curr][j].second;
      }
    }
    cout << sum << " " << minn << '\n';
  }
  return 0;
}