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 OO = (int)2e9;
const double eps = 1e-9;
int daysInMonths[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int di[] = {-1, 0, 1, 0};
int dj[] = {0, 1, 0, -1};
long long k, j, w, dp[100002][35], sum[100002][35], mn[100002][35];
int out[100002], weight[100002], n;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  scanf("%d%I64d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &out[i]);
    dp[i][0] = out[i];
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &weight[i]);
    sum[i][0] = weight[i];
    mn[i][0] = weight[i];
  }
  for (int k = 1; k <= 34; k++) {
    for (int u = 0; u < n; u++) {
      dp[u][k] = dp[dp[u][k - 1]][k - 1];
      sum[u][k] = sum[u][k - 1] + sum[dp[u][k - 1]][k - 1];
      mn[u][k] = min(mn[u][k - 1], mn[dp[u][k - 1]][k - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    long long s = 0;
    long long m = 1e18;
    long long rem = k;
    int u = i;
    for (int k = 34; k >= 0; k--) {
      if (rem >= (1LL << k)) {
        s += sum[u][k];
        m = min(m, mn[u][k]);
        rem -= (1LL << k);
        u = dp[u][k];
      }
    }
    printf("%I64d %d\n", s, m);
  }
  return 0;
}
