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 n;
long long k;
long long f[34][100001];
int g[34][100001];
int h[34][100001];
int main() {
  scanf("%d%lld", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &g[0][i]), g[0][i]++;
  for (int i = 1; i <= n; i++) scanf("%d", &h[0][i]), f[0][i] = h[0][i];
  for (int i = 1; i <= 33; i++)
    for (int j = 1; j <= n; j++) {
      f[i][j] = f[i - 1][j] + f[i - 1][g[i - 1][j]];
      g[i][j] = g[i - 1][g[i - 1][j]];
      h[i][j] = min(h[i - 1][j], h[i - 1][g[i - 1][j]]);
    }
  for (int i = 1; i <= n; i++) {
    int now = i;
    long long ans1 = 0;
    int ans2 = 1000000001;
    for (int j = 33; j >= 0; j--)
      if (k >> j & 1)
        ans1 += f[j][now], ans2 = min(ans2, h[j][now]), now = g[j][now];
    printf("%lld %d\n", ans1, ans2);
  }
}
