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 int LIM = 1e+5 + 100;
const int MAX_PW = 34;
int f[LIM];
int w[LIM];
int _min[LIM][MAX_PW];
long long sum[LIM][MAX_PW];
int to[LIM][MAX_PW];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  long long k;
  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][0] = _min[i][0] = w[i];
    to[i][0] = f[i];
  }
  for (int pw = 1; pw < MAX_PW; ++pw) {
    for (int i = 0; i < LIM; ++i) {
      _min[i][pw] = min(_min[i][pw - 1], _min[to[i][pw - 1]][pw - 1]);
      sum[i][pw] = sum[i][pw - 1] + sum[to[i][pw - 1]][pw - 1];
      to[i][pw] = to[to[i][pw - 1]][pw - 1];
    }
  }
  for (int i = 0; i < n; ++i) {
    int min_ans = 1e+9;
    long long sum_ans = 0;
    int cur = i;
    for (int pw = MAX_PW; pw >= 0; --pw)
      if ((k >> pw) & 1) {
        min_ans = min(min_ans, _min[cur][pw]);
        sum_ans += sum[cur][pw];
        cur = to[cur][pw];
      }
    cout << sum_ans << " " << min_ans << '\n';
  }
  return 0;
}
