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 maxn = 1e5 + 7;
int fa[maxn][35];
long long minval[maxn][35];
long long sum[maxn][35];
void init(int n) {
  for (int k = 1; k < 34; k++)
    for (int i = 0; i < n; i++) {
      fa[i][k] = fa[fa[i][k - 1]][k - 1];
      minval[i][k] = min(minval[i][k - 1], minval[fa[i][k - 1]][k - 1]);
      sum[i][k] = sum[i][k - 1] + sum[fa[i][k - 1]][k - 1];
    }
}
void solve(int r, long long m) {
  long long ansmin = 1e9 + 6;
  long long anssum = 0;
  for (int k = 0; k < 34; k++)
    if (m & (1LL << k)) {
      anssum += sum[r][k];
      ansmin = min(ansmin, minval[r][k]);
      r = fa[r][k];
    }
  printf("%lld %lld\n", anssum, ansmin);
}
int main() {
  int n;
  long long m;
  scanf("%d%lld", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &fa[i][0]);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &minval[i][0]);
    sum[i][0] = minval[i][0];
  }
  init(n);
  for (int i = 0; i < n; i++) solve(i, m);
}