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;
void test_case() {
  long long n, i, j, k;
  cin >> n >> k;
  vector<long long> p(n);
  vector<long long> w(n);
  const long long LOGN = log2(k) + 5;
  for (i = 0; i < n; i++) {
    cin >> p[i];
  }
  for (i = 0; i < n; i++) {
    cin >> w[i];
  }
  vector<vector<long long>> sparse(n, vector<long long>(LOGN, -1));
  vector<vector<pair<long long, long long>>> sparse2(
      n, vector<pair<long long, long long>>(LOGN, {0, INT_MAX}));
  for (i = 0; i < n; i++) {
    sparse[i][0] = p[i];
    sparse2[i][0] = {w[i], w[i]};
  }
  for (j = 1; j < LOGN; j++) {
    for (i = 0; i < n; i++) {
      assert(sparse[i][j - 1] != -1);
      sparse[i][j] = sparse[sparse[i][j - 1]][j - 1];
      sparse2[i][j].first =
          sparse2[i][j - 1].first + sparse2[sparse[i][j - 1]][j - 1].first;
      sparse2[i][j].second = min(sparse2[i][j - 1].second,
                                 sparse2[sparse[i][j - 1]][j - 1].second);
    }
  }
  for (i = 0; i < n; i++) {
    long long cnt = k;
    long long su = 0;
    long long mn = INT_MAX;
    long long u = i;
    for (j = LOGN - 1; j >= 0; j--) {
      if (cnt - (1LL << j) >= 0) {
        cnt -= (1LL << j);
        su += sparse2[u][j].first;
        mn = min(mn, sparse2[u][j].second);
        u = sparse[u][j];
      }
    }
    assert(cnt == 0);
    cout << su << " " << mn << "\n";
  }
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  t = 1;
  while (t--) {
    test_case();
  }
  return 0;
}