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;
long long i, i1, j, k, k1, t, n, m, res, flag[10], a, b, s;
long long f[100010], w[100010], jm[35][100010], sts[35][100010],
    stm[35][100010];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> f[i];
    f[i]++;
  }
  for (i = 1; i <= n; i++) {
    cin >> w[i];
  }
  for (j = 1; j <= n; j++) {
    jm[0][j] = f[j];
    sts[0][j] = w[j];
    stm[0][j] = w[j];
  }
  for (i = 1; i < 35; i++) {
    for (j = 1; j <= n; j++) {
      jm[i][j] = jm[i - 1][jm[i - 1][j]];
      sts[i][j] = sts[i - 1][j] + sts[i - 1][jm[i - 1][j]];
      stm[i][j] = min(stm[i - 1][j], stm[i - 1][jm[i - 1][j]]);
    }
  }
  for (i = 1; i <= n; i++) {
    a = 0;
    b = (long long)1e18;
    s = i;
    for (j = 0; j < 35; j++) {
      if ((k >> j) & 1) {
        a += sts[j][s];
        b = min(b, stm[j][s]);
        s = jm[j][s];
      }
    }
    cout << a << ' ' << b << "\n";
  }
  return 0;
}
