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;
const int N = 100000;
const int M = 34;
int to[M][N + 5];
int mn[M][N + 5];
long long sum[M][N + 5];
int T1[N + 5], T2[N + 5];
int M1[N + 5], M2[N + 5];
long long S1[N + 5], S2[N + 5];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> to[0][i];
  for (int i = 0; i < n; i++) {
    cin >> sum[0][i];
    mn[0][i] = sum[0][i];
  }
  for (int i = 1; i < M; i++) {
    for (int j = 0; j < n; j++) {
      to[i][j] = to[i - 1][to[i - 1][j]];
      mn[i][j] = min(mn[i - 1][j], mn[i - 1][to[i - 1][j]]);
      sum[i][j] = sum[i - 1][j] + sum[i - 1][to[i - 1][j]];
    }
  }
  vector<int> B;
  for (int i = 0; i < M; i++)
    if ((k & (1LL << i)) != 0) B.push_back(i);
  for (int i = 0; i < n; i++) {
    T1[i] = to[B[0]][i];
    M1[i] = mn[B[0]][i];
    S1[i] = sum[B[0]][i];
  }
  for (int i = 1; i < B.size(); i++) {
    for (int j = 0; j < n; j++) {
      T2[j] = to[B[i]][T1[j]];
      M2[j] = min(M1[j], mn[B[i]][T1[j]]);
      S2[j] = S1[j] + sum[B[i]][T1[j]];
    }
    for (int j = 0; j < n; j++) {
      T1[j] = T2[j];
      M1[j] = M2[j];
      S1[j] = S2[j];
    }
  }
  for (int i = 0; i < n; i++) cout << S1[i] << " " << M1[i] << "\n";
  return 0;
}
