Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int n, dp[N][45];
long long minn[N][45], sum[N][45];
long long k;
void solve(int xx) {
  int x = xx;
  long long ret = 0;
  for (int i = 0; i <= 40; ++i) {
    if ((k >> i) & 1) {
      ret += sum[x][i];
      x = dp[x][i];
    }
  }
  printf("%lld ", ret);
  ret = 1000000000000;
  x = xx;
  for (int i = 0; i <= 40; ++i) {
    if ((k >> i) & 1) {
      ret = min(ret, minn[x][i]);
      x = dp[x][i];
    }
  }
  printf("%lld\n", ret);
}
int main() {
  scanf("%d%lld", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &dp[i][0]);
    ++dp[i][0];
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &sum[i][0]);
    minn[i][0] = sum[i][0];
  }
  for (int j = 1; j <= 40; ++j) {
    for (int i = 1; i <= n; ++i) {
      dp[i][j] = dp[dp[i][j - 1]][j - 1];
    }
  }
  for (int j = 1; j <= 40; ++j) {
    for (int i = 1; i <= n; ++i) {
      sum[i][j] = sum[i][j - 1] + sum[dp[i][j - 1]][j - 1];
    }
  }
  for (int j = 1; j <= 40; ++j) {
    for (int i = 1; i <= n; ++i) {
      minn[i][j] = min(minn[i][j - 1], minn[dp[i][j - 1]][j - 1]);
    }
  }
  for (int i = 1; i <= n; ++i) solve(i);
  return 0;
}