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 f[100005][40], g[100005][40], h[100005][40];
long long n, m, x, ans1, ans2;
inline long long read() {
  long long x = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - 48;
  return x;
}
int main() {
  n = read();
  m = read();
  for (long long i = 0; i < n; i++) f[i][0] = read();
  for (long long i = 0; i < n; i++) g[i][0] = h[i][0] = read();
  for (long long i = 1; i <= 35; i++)
    for (long long j = 0; j < n; j++) {
      f[j][i] = f[f[j][i - 1]][i - 1];
      g[j][i] = g[j][i - 1] + g[f[j][i - 1]][i - 1];
      h[j][i] = min(h[j][i - 1], h[f[j][i - 1]][i - 1]);
    }
  for (long long i = 0; i < n; i++) {
    ans2 = 1e10;
    ans1 = 0;
    x = i;
    for (long long tmp = m, p = 0; tmp; tmp /= 2, p++)
      if (tmp % 2) {
        ans1 += g[x][p];
        ans2 = min(ans2, h[x][p]);
        x = f[x][p];
      }
    printf("%I64d %I64d\n", ans1, ans2);
  }
}
