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 = 1e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n;
long long k;
long long mi[N][50];
long long ans[N][50];
long long f[N][50];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  int i, j;
  for (i = 0; i < n; i++) {
    cin >> f[i][0];
  }
  for (i = 0; i < n; i++) {
    cin >> ans[i][0];
    mi[i][0] = ans[i][0];
  }
  for (long long i = 1; i < 50; ++i) {
    for (long long j = 0; j < n; ++j) {
      long long nxt = f[j][i - 1];
      f[j][i] = f[nxt][i - 1];
      ans[j][i] = ans[j][i - 1] + ans[nxt][i - 1];
      mi[j][i] = min(mi[j][i - 1], mi[nxt][i - 1]);
    }
  }
  for (long long u = 0; u < n; ++u) {
    long long s = 0, mx = 1e18, pos = u, x = k;
    for (long long i = 49; i >= 0; i--)
      if ((x >> i) & 1) {
        s += ans[pos][i];
        mx = min(mx, mi[pos][i]);
        pos = f[pos][i];
      }
    cout << s << " " << mx << endl;
  }
  return 0;
}