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 + 7;
const int M = 2e6 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
long long w[N], k;
int f[N], nx[N][35], n;
long long sum[N][35], mn[N][35];
int main() {
  scanf("%d", &n);
  scanf("%lld", &k);
  for (int i = 0; i < n; i++) scanf("%d", &f[i]);
  for (int i = 0; i < n; i++) scanf("%lld", &w[i]);
  for (int i = 0; i < n; i++)
    nx[i][0] = f[i], sum[i][0] = w[i], mn[i][0] = w[i];
  for (int j = 1; j < 35; j++) {
    for (int i = 0; i < n; i++) {
      nx[i][j] = nx[nx[i][j - 1]][j - 1];
      mn[i][j] = min(mn[i][j - 1], mn[nx[i][j - 1]][j - 1]);
      sum[i][j] = sum[i][j - 1] + sum[nx[i][j - 1]][j - 1];
    }
  }
  for (int i = 0; i < n; i++) {
    int now = i;
    long long ans = 0, ret = k, minn = INF;
    for (int j = 34; j >= 0; j--) {
      if (ret - (1ll << j) >= 0) {
        ret -= (1ll << j);
        ans += sum[now][j];
        minn = min(minn, mn[now][j]);
        now = nx[now][j];
      }
    }
    printf("%lld %lld\n", ans, minn);
  }
  return 0;
}
