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 N = 1e5 + 5;
int nxt[N][36];
long long sum[N][36];
int mi[N][36];
int main() {
  int n;
  long long k;
  while (cin >> n >> k) {
    for (int i = 0; i < n; i++) {
      int x;
      scanf("%d", &x);
      nxt[i][0] = x;
    }
    for (int i = 0; i < n; i++) {
      int x;
      scanf("%d", &x);
      sum[i][0] = x;
      mi[i][0] = x;
    }
    for (int j = 1; j <= 35; j++) {
      for (int i = 0; i < n; i++) {
        nxt[i][j] = nxt[nxt[i][j - 1]][j - 1];
        mi[i][j] = min(mi[i][j - 1], mi[nxt[i][j - 1]][j - 1]);
        sum[i][j] = sum[i][j - 1] + sum[nxt[i][j - 1]][j - 1];
      }
    }
    for (int i = 0; i < n; i++) {
      long long ans1 = 0;
      int ans2 = 1e9;
      int cur = i;
      for (int j = 0; j <= 35; j++) {
        if (k & ((long long)1 << j)) {
          ans1 += sum[cur][j];
          ans2 = min(ans2, mi[cur][j]);
          cur = nxt[cur][j];
        }
      }
      cout << ans1 << " " << ans2 << endl;
    }
  }
  return 0;
}
