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;
int N;
long long K;
int f[1 << 18];
long long c[1 << 18];
int j[1 << 18][35];
long long jc[1 << 18][35];
long long jmin[1 << 18][35];
long long ret[1 << 18][2];
int main() {
  scanf("%d %lld", &N, &K);
  for (int i = 0; i < N; i++) {
    scanf("%d", &f[i]);
    j[i][0] = f[i];
  }
  for (int i = 0; i < N; i++) {
    scanf("%lld", &c[i]);
    jc[i][0] = c[i];
    jmin[i][0] = c[i];
  }
  for (int k = 1; k < 35; k++) {
    for (int i = 0; i < N; i++) {
      int mid = j[i][k - 1];
      j[i][k] = j[mid][k - 1];
      jc[i][k] = jc[i][k - 1] + jc[mid][k - 1];
      jmin[i][k] = min(jmin[i][k - 1], jmin[mid][k - 1]);
    }
  }
  for (int i = 0; i < N; i++) {
    int u = i;
    long long cost = 0;
    long long mincost = 2e10;
    for (int k = 0; k < 35; k++) {
      if ((K & (1ULL << k)) != 0) {
        cost += jc[u][k];
        mincost = min(mincost, jmin[u][k]);
        u = j[u][k];
      }
    }
    ret[i][0] = cost;
    ret[i][1] = mincost;
  }
  for (int i = 0; i < N; i++) {
    printf("%lld %lld\n", ret[i][0], ret[i][1]);
  }
}
