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 k;
int n;
struct ans {
  long long sum;
  int m;
  int dest;
};
ans a[40][100005];
int main() {
  scanf("%d %lld", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[0][i].dest);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[0][i].m);
    a[0][i].sum = a[0][i].m;
  }
  for (int i = 1; i <= 39; i++) {
    for (int j = 0; j < n; j++) {
      a[i][j].dest = a[i - 1][a[i - 1][j].dest].dest;
      a[i][j].sum = a[i - 1][j].sum + a[i - 1][a[i - 1][j].dest].sum;
      a[i][j].m = min(a[i - 1][j].m, a[i - 1][a[i - 1][j].dest].m);
    }
  }
  for (int i = 0; i < n; i++) {
    int dest = i;
    long long sum = 0;
    int m = 199999999;
    for (int j = 0; j <= 39; j++) {
      if (!((1LL << j) & k)) continue;
      sum += a[j][dest].sum;
      m = min(m, a[j][dest].m);
      dest = a[j][dest].dest;
    }
    printf("%lld %d\n", sum, m);
  }
  return 0;
}
