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;
struct V {
  int p[35];
  long long s[35];
  long long m[35];
} v[100000];
int main() {
  int n;
  long long k;
  scanf("%d%I64d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &v[i].p[0]);
  for (int i = 0; i < n; i++) {
    int w;
    scanf("%d", &w);
    v[i].m[0] = v[i].s[0] = w;
  }
  for (int j = 1; j < 35; j++) {
    for (int i = 0; i < n; i++) {
      V& cv = v[i];
      cv.p[j] = v[cv.p[j - 1]].p[j - 1];
      cv.s[j] = cv.s[j - 1] + v[cv.p[j - 1]].s[j - 1];
      cv.m[j] = min(cv.m[j - 1], v[cv.p[j - 1]].m[j - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    int cur = i;
    long long m = 0x1234567890123456LL;
    long long s = 0;
    for (int j = 0; j < 35; j++) {
      if ((k & (1LL << j)) == 0) continue;
      m = min(m, v[cur].m[j]);
      s += v[cur].s[j];
      cur = v[cur].p[j];
    }
    printf("%I64d %I64d\n", s, m);
  }
  return 0;
}
