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;
long long int n, k;
const int N = 100 * 1000;
long long int f[N], w[N], sum[N], mini[N], to[N], tsum[N], tmini[N], tto[N];
void sol(long long int p) {
  if (p == 1) return;
  sol(p / 2);
  for (int i = 0; i < n; i++) {
    tsum[i] = sum[i];
    tmini[i] = mini[i];
    tto[i] = to[i];
  }
  for (int i = 0; i < n; i++) {
    sum[i] += tsum[tto[i]];
    mini[i] = min(tmini[i], tmini[tto[i]]);
    to[i] = tto[tto[i]];
  }
  if (p % 2 == 1) {
    for (int i = 0; i < n; i++) {
      sum[i] += w[to[i]];
      mini[i] = min(mini[i], w[to[i]]);
      to[i] = f[to[i]];
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> f[i];
  for (int i = 0; i < n; i++) cin >> w[i];
  for (int i = 0; i < n; i++) {
    sum[i] = w[i];
    mini[i] = w[i];
    to[i] = f[i];
  }
  sol(k);
  for (int i = 0; i < n; i++) cout << sum[i] << ' ' << mini[i] << '\n';
  return 0;
}