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>
#pragma GCC optimize("Ofast")
using namespace std;
clock_t clk = clock();
long long int i, j, k;
void solve(void);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t = 1;
  while (t--) solve();
  return 0;
}
long long int n;
long long int f[100005], w[100005];
long long int dp[50][100005][3];
void solve() {
  cin >> n >> k;
  for (i = 0; i <= n - 1; ++i) cin >> f[i];
  for (i = 0; i <= n - 1; ++i) cin >> w[i];
  for (i = 0; i <= n - 1; ++i) {
    dp[0][i][0] = f[i];
    dp[0][i][1] = w[i];
    dp[0][i][2] = w[i];
  }
  for (j = 1; j <= 35; ++j)
    for (i = 0; i <= n - 1; ++i) {
      long long int next = dp[j - 1][i][0];
      dp[j][i][0] = dp[j - 1][next][0];
      dp[j][i][1] = dp[j - 1][i][1] + dp[j - 1][next][1];
      dp[j][i][2] =
          ((dp[j - 1][i][2]) < (dp[j - 1][next][2]) ? (dp[j - 1][i][2])
                                                    : (dp[j - 1][next][2]));
    }
  vector<long long int> v;
  while (k > 0) {
    long long int t = log2(k);
    v.push_back(t);
    k -= ((long long int)1 << t);
  }
  for (i = 0; i <= n - 1; ++i) {
    long long int next = dp[v[0]][i][0];
    long long int sum = dp[v[0]][i][1];
    long long int min = dp[v[0]][i][2];
    for (j = 1; j <= (int)v.size() - 1; ++j) {
      sum += dp[v[j]][next][1];
      min = ((min) < (dp[v[j]][next][2]) ? (min) : (dp[v[j]][next][2]));
      next = dp[v[j]][next][0];
    }
    cout << sum << " " << min << '\n';
  }
}