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 f[44][100050], g[44][100050], n;
long long h[44][100050], K;
int main() {
  scanf("%d%lld", &n, &K);
  int i, x, j;
  for (i = 1; i <= n; i++) {
    scanf("%d", &f[0][i]);
    f[0][i]++;
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    g[0][i] = h[0][i] = x;
  }
  for (i = 1; (1ll << i) <= K; i++) {
    for (j = 1; j <= n; j++) {
      f[i][j] = f[i - 1][f[i - 1][j]];
      g[i][j] = min(g[i - 1][j], g[i - 1][f[i - 1][j]]);
      h[i][j] = h[i - 1][j] + h[i - 1][f[i - 1][j]];
    }
  }
  for (i = 1; i <= n; i++) {
    long long t = K;
    int x = i;
    int ans1 = 1 << 30;
    long long ans2 = 0;
    for (j = 0; j <= 40; j++) {
      if (t & (1ll << j)) {
        ans1 = min(ans1, g[j][x]);
        ans2 += h[j][x];
        x = f[j][x];
      }
    }
    printf("%lld %d\n", ans2, ans1);
  }
}
