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(100005);
const long long inf((long long)1 << 60);
int n;
long long k, minn[35], ver[N][35], edge1[N][35], edge2[N][35];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i(1); i <= n; ++i) cin >> ver[i][0], ++ver[i][0];
  for (int i(1); i <= n; ++i) {
    cin >> edge1[i][0];
    edge2[i][0] = edge1[i][0];
  }
  for (int i(1); i <= 34; ++i) {
    for (int j(1); j <= n; ++j) {
      ver[j][i] = ver[ver[j][i - 1]][i - 1];
      edge1[j][i] = edge1[ver[j][i - 1]][i - 1] + edge1[j][i - 1];
      edge2[j][i] = ((edge2[ver[j][i - 1]][i - 1]) < (edge2[j][i - 1])
                         ? (edge2[ver[j][i - 1]][i - 1])
                         : (edge2[j][i - 1]));
    }
  }
  minn[0] = 1;
  for (int i(1); i <= 34; ++i) minn[i] = minn[i - 1] << 1;
  for (int i(1); i <= n; ++i) {
    int now(i);
    long long step(k), ans1(0), ans2(inf);
    for (int i(34); i >= 0; --i) {
      if (step >= minn[i]) {
        step -= minn[i];
        ans1 += edge1[now][i];
        ans2 = ((ans2) < (edge2[now][i]) ? (ans2) : (edge2[now][i]));
        now = ver[now][i];
      }
    }
    cout << ans1 << " " << ans2 << "\n";
  }
  return 0;
}