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 up[100010][35], upsum[100010][35], upmin[100010][35];
long long n, k;
void build() {
  for (int k = 1; k <= 34; k++)
    for (int j = 0; j < n; j++) up[j][k] = up[up[j][k - 1]][k - 1];
  for (int k = 1; k <= 34; k++)
    for (int j = 0; j < n; j++)
      upsum[j][k] = upsum[up[j][k - 1]][k - 1] + upsum[j][k - 1];
  for (int k = 1; k <= 34; k++)
    for (int j = 0; j < n; j++)
      upmin[j][k] = min(upmin[up[j][k - 1]][k - 1], upmin[j][k - 1]);
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0, a; i < n; i++) {
    cin >> a;
    up[i][0] = a;
  }
  for (int i = 0, a; i < n; i++) {
    cin >> a;
    upmin[i][0] = upsum[i][0] = a;
  }
  build();
  for (int i = 0; i < n; i++) {
    long long x = i;
    long long sum = 0, minn = 1e9;
    for (int j = 34; j >= 0; j--) {
      if ((1LL << j) & k) {
        sum += upsum[x][j];
        minn = min(minn, upmin[x][j]);
        x = up[x][j];
      }
    }
    cout << sum << " " << minn << '\n';
  }
  return 0;
}