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 long long N = 100010, K = 50;
long long n, k, son[N][K], mis[N][K], mi[N][K];
long long getMis(long long c, long long x) {
  if (!x) return x;
  long long p = 0, dp = 1;
  while (dp <= x) {
    p++;
    dp <<= 1;
  }
  p--;
  dp >>= 1;
  return mis[c][p] + getMis(son[c][p], x - dp);
}
long long getMi(long long c, long long x) {
  if (!x) return 0x3f3f3f3f;
  long long p = 0, dp = 1;
  while (dp <= x) {
    p++;
    dp <<= 1;
  }
  p--;
  dp >>= 1;
  return min(mi[c][p], getMi(son[c][p], x - dp));
}
int main() {
  cin >> n >> k;
  for (long long i = 0; i < n; i++) cin >> son[i][0];
  for (long long i = 0; i < n; i++) {
    cin >> mis[i][0];
    mi[i][0] = mis[i][0];
  }
  for (long long p = 1; p < K; p++) {
    for (long long i = 0; i < n; i++) {
      son[i][p] = son[son[i][p - 1]][p - 1];
    }
  }
  for (long long p = 1; p < K; p++) {
    for (long long i = 0; i < n; i++) {
      mi[i][p] = min(mi[i][p - 1], mi[son[i][p - 1]][p - 1]);
      mis[i][p] = mis[i][p - 1] + mis[son[i][p - 1]][p - 1];
    }
  }
  for (long long i = 0; i < n; i++) {
    cout << getMis(i, k) << ' ' << getMi(i, k) << endl;
  }
  return 0;
}
