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 inf = 1e9 + 1;
const double eps = 1e-9;
const int MAXN = 1e5 + 1;
const int LOG = 34;
int n;
long long k;
int m[MAXN], w[MAXN];
int lca[LOG][MAXN];
long long sum[LOG][MAXN];
int mi[LOG][MAXN];
pair<long long, int> f(int a) {
  long long l = k;
  pair<long long, int> ret;
  ret.second = inf;
  for (int i = LOG - 1; i >= 0; i--) {
    if (((long long)1 << i) <= (long long)l) {
      ret.first = (long long)ret.first + (long long)sum[i][a];
      ret.second = min(ret.second, mi[i][a]);
      a = lca[i][a];
      l -= ((long long)1 << i);
    }
  }
  return ret;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> m[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> w[i];
    lca[0][i] = m[i];
    sum[0][i] = w[i];
    mi[0][i] = w[i];
  }
  for (int i = 1; i < LOG; i++) {
    for (int j = 0; j < n; j++) {
      lca[i][j] = lca[i - 1][lca[i - 1][j]];
      sum[i][j] =
          (long long)sum[i - 1][j] + (long long)sum[i - 1][lca[i - 1][j]];
      mi[i][j] = min(mi[i - 1][j], mi[i - 1][lca[i - 1][j]]);
    }
  }
  for (int i = 0; i < n; i++) {
    pair<long long, int> curr = f(i);
    cout << curr.first << " " << curr.second << '\n';
  }
  return 0;
}
