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
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e5 + 5;
const long long mod = 1e9 + 7;
long long n, k;
struct node {
  long long to, sum, mn;
};
node a[100010], ans[100010];
void solve() {
  while (k) {
    if (k & 1) {
      vector<node> tmp(n);
      for (long long i = 0; i <= n - 1; ++i) {
        tmp[i] = {a[ans[i].to].to, a[ans[i].to].sum + ans[i].sum,
                  min(a[ans[i].to].mn, ans[i].mn)};
      }
      for (long long i = 0; i <= n - 1; ++i) ans[i] = tmp[i];
    }
    vector<node> tmp(n);
    for (long long i = 0; i <= n - 1; ++i) {
      tmp[i] = {a[a[i].to].to, a[a[i].to].sum + a[i].sum,
                min(a[a[i].to].mn, a[i].mn)};
    }
    for (long long i = 0; i <= n - 1; ++i) a[i] = tmp[i];
    k >>= 1;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = 0; i <= n - 1; ++i) cin >> a[i].to;
  for (long long i = 0; i <= n - 1; ++i) {
    cin >> a[i].mn;
    a[i].sum = a[i].mn;
    ans[i] = a[i];
  }
  k--;
  solve();
  for (long long i = 0; i <= n - 1; ++i)
    cout << ans[i].sum << "  " << ans[i].mn << '\n';
  return 0;
}
