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;
int N;
long long K;
int to[100005];
long long W[100005];
int pa[100005][35];
long long st1[100005][35], st2[100005][35];
long long ans1[100005], ans2[100005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> K;
  for (int i = 0; i <= N - 1; i++) cin >> to[i];
  for (int i = 0; i <= N - 1; i++) {
    cin >> W[i];
    pa[i][0] = to[i];
    st1[i][0] = W[i];
    st2[i][0] = W[i];
  }
  long long ln = 2;
  int lg = 1;
  int mxlg = 0;
  while (ln <= K) {
    mxlg++;
    for (int i = 0; i <= N - 1; i++) {
      pa[i][lg] = pa[pa[i][lg - 1]][lg - 1];
      st1[i][lg] = st1[i][lg - 1] + st1[pa[i][lg - 1]][lg - 1];
      st2[i][lg] = min(st2[i][lg - 1], st2[pa[i][lg - 1]][lg - 1]);
    }
    ln <<= 1;
    lg++;
  }
  for (int i = 0; i <= N - 1; i++) {
    ans1[i] = 0;
    ans2[i] = (1LL << 60);
    int u = i;
    for (int b = mxlg; b >= 0; b--) {
      if (K & (1LL << b)) {
        ans1[i] += st1[u][b];
        ans2[i] = min(ans2[i], st2[u][b]);
        u = pa[u][b];
      }
    }
  }
  for (int i = 0; i <= N - 1; i++) {
    cout << ans1[i] << ' ' << ans2[i] << '\n';
  }
  return 0;
}
