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;
int m[100005][34], go[100005][34];
long long sum[100005][34];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<int> next(n), w(n);
  for (auto &i : next) cin >> i;
  for (auto &i : w) cin >> i;
  for (int u = 0; u < n; ++u) {
    m[u][0] = w[u];
    sum[u][0] = w[u];
    go[u][0] = next[u];
  }
  for (int i = 1; i < 34; ++i)
    for (int u = 0; u < n; ++u) {
      int v = go[u][i - 1];
      m[u][i] = min(m[u][i - 1], m[v][i - 1]);
      sum[u][i] = sum[u][i - 1] + sum[v][i - 1];
      go[u][i] = go[v][i - 1];
    }
  for (int i = 0; i < n; ++i) {
    long long s = 0, tk = k;
    int mn = numeric_limits<int>::max(), u = i;
    for (int j = 33; j >= 0; --j)
      if (tk - (1LL << j) >= 0) {
        tk -= 1LL << j;
        s += sum[u][j];
        mn = min(mn, m[u][j]);
        u = go[u][j];
      }
    cout << s << " " << mn << "\n";
  }
  return 0;
}
