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>
const long long INF = 4e18L + 1;
const int IINF = 2e9 + 1;
const int limit = 1048576;
using namespace std;
const long long LOG = 35;
vector<vector<long long> > s_val, m_val, next_vert;
long long s, m;
void solve(long long v, long long len, long long scanned) {
  for (long long i = (long long)scanned; i < (long long)LOG; i++) {
    if (len & (1ll << i)) {
      if (0) cout << "i " << i << " len " << len << " s " << s << endl;
      s += s_val[v][i];
      m = min(m, m_val[v][i]);
      solve(next_vert[v][i], len, i + 1);
      return;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, k;
  cin >> n >> k;
  s_val.resize(n, vector<long long>(LOG)),
      m_val.resize(n, vector<long long>(LOG)),
      next_vert.resize(n, vector<long long>(LOG));
  for (auto& arr : next_vert) {
    cin >> arr[0];
  }
  for (auto& arr : s_val) {
    cin >> arr[0];
  }
  m_val = s_val;
  for (long long i = (long long)1; i < (long long)LOG; i++) {
    for (long long v = (long long)0; v < (long long)n; v++) {
      int w = next_vert[v][i - 1];
      next_vert[v][i] = next_vert[w][i - 1];
      s_val[v][i] = s_val[v][i - 1] + s_val[w][i - 1];
      m_val[v][i] = min(m_val[v][i - 1], m_val[w][i - 1]);
      if (0)
        cout << "v " << v << " i " << i << " s_val " << s_val[v][i] << endl;
    }
  }
  for (long long v = (long long)0; v < (long long)n; v++) {
    s = 0, m = INF;
    solve(v, k, 0);
    cout << s << " " << m << "\n";
  }
  return 0;
}