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 INF = 2e9;
const long long INFL = 2e18;
const int N = 1e5, LOG = 35;
const long long K = 1e10;
int n;
long long k;
int m[N + 5][LOG + 5], p[N + 5][LOG + 5];
long long s[N + 5][LOG + 5];
void qr(int u, long long k) {
  int Min = 1e9;
  long long S = 0;
  for (int i = (LOG), _b = (0); i >= _b; --i)
    if ((k - (1LL << i)) >= 0) {
      k -= (1LL << i);
      S += s[u][i];
      Min = min(Min, m[u][i]);
      u = p[u][i];
    }
  cout << S << ' ' << Min << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = (1), _b = (n); i <= _b; ++i) {
    int tmp;
    cin >> tmp;
    p[i][0] = tmp + 1;
  }
  for (int i = (1), _b = (n); i <= _b; ++i) {
    int tmp;
    cin >> tmp;
    m[i][0] = s[i][0] = tmp;
  }
  for (int j = (1), _b = (LOG); j <= _b; ++j)
    for (int i = (1), _b = (n); i <= _b; ++i) {
      p[i][j] = p[p[i][j - 1]][j - 1];
      s[i][j] = s[i][j - 1] + s[p[i][j - 1]][j - 1];
      m[i][j] = min(m[i][j - 1], m[p[i][j - 1]][j - 1]);
    }
  for (int i = (1), _b = (n); i <= _b; ++i) qr(i, k);
  return 0;
}
