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;
long double eps = 1e-7;
const int N = 1e5 + 10;
long long dp[N][36], mn[N][36], sum[N][36];
long long n, k;
pair<long long, long long> mnAndSum(int node) {
  long long sum1 = 0, mn1 = 2e18, k1 = k;
  for (long long i = 34; i >= 0; i--)
    if ((1LL << i) <= k1)
      k1 -= (1LL << i), mn1 = min(mn1, mn[node][i]), sum1 += sum[node][i],
                        node = dp[node][i];
  return {sum1, mn1};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  memset(dp, -1, sizeof dp);
  for (long long i = 0; i < n; i++) {
    long long next;
    cin >> next;
    dp[i][0] = next;
  }
  for (long long i = 0; i < n; i++) {
    long long cost;
    cin >> cost;
    mn[i][0] = cost;
    sum[i][0] = cost;
  }
  for (long long i = 1; i <= 34; i++) {
    for (long long j = 0; j < n; j++) {
      if (dp[j][i - 1] != -1) {
        dp[j][i] = dp[dp[j][i - 1]][i - 1];
        mn[j][i] = min(mn[j][i - 1], mn[dp[j][i - 1]][i - 1]);
        sum[j][i] = sum[j][i - 1] + sum[dp[j][i - 1]][i - 1];
      }
    }
  }
  for (long long i = 0; i < n; i++) {
    pair<long long, long long> ans = mnAndSum(i);
    cout << ans.first << " " << ans.second << '\n';
  }
}
