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;
struct step {
  long long to, len, min;
  step(){};
  step(pair<long, long> p) {
    to = p.first;
    len = p.second;
    min = p.second;
  }
};
long long n, k, l = 0;
vector<bool> used;
vector<pair<long long, long long> > g, ans;
vector<vector<step> > f;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  g.resize(n);
  ans.resize(n);
  used.resize(n, false);
  for (long long i = 0; i < n; i++) cin >> g[i].first;
  for (long long i = 0; i < n; i++) cin >> g[i].second;
  while ((1ll << l) <= k) l++;
  f.resize(l, vector<step>(n));
  for (long long i = 0; i < l; i++) {
    for (long long y = 0; y < n; y++) {
      if (i == 0)
        f[i][y] = step(g[y]);
      else {
        f[i][y].to = f[i - 1][f[i - 1][y].to].to;
        f[i][y].len = f[i - 1][y].len + f[i - 1][f[i - 1][y].to].len;
        f[i][y].min = min(f[i - 1][y].min, f[i - 1][f[i - 1][y].to].min);
      }
    }
  }
  for (long long i = 0; i < n; i++) {
    long long kk = k, from = i, len = 0, mymin = 1e18, it = l;
    while (kk > 0) {
      if ((1ll << it) <= kk) {
        kk -= (1ll << it);
        len += f[it][from].len;
        mymin = min(mymin, f[it][from].min);
        from = f[it][from].to;
      } else
        it--;
    }
    ans[i] = {len, mymin};
  }
  for (long long i = 0; i < n; i++)
    cout << ans[i].first << ' ' << ans[i].second << '\n';
  return 0;
}
