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 = 100005;
int child[N], w[N];
long long dp[N][35][3], ans[N][2];
void pre(int n) {
  for (int i = 0; i < n; i++) {
    dp[i][0][0] = w[i];
    dp[i][0][2] = w[i];
    dp[i][0][1] = child[i];
  }
  for (int j = 1; j < 35; j++) {
    for (int i = 0; i < n; i++) {
      dp[i][j][1] = dp[dp[i][j - 1][1]][j - 1][1];
      dp[i][j][0] = dp[i][j - 1][0] + dp[dp[i][j - 1][1]][j - 1][0];
      dp[i][j][2] = min(dp[i][j - 1][2], dp[dp[i][j - 1][1]][j - 1][2]);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> child[i];
  for (int i = 0; i < n; i++) cin >> w[i];
  pre(n);
  int pt;
  for (int i = 0; i < n; i++) {
    pt = i;
    ans[i][1] = 1000000007ULL;
    for (long long j = 0; (1LL << j) <= k; j++) {
      if ((1LL << j) & k) {
        ans[i][0] += dp[pt][j][0];
        ans[i][1] = min(ans[i][1], dp[pt][j][2]);
        pt = dp[pt][j][1];
      }
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i][0] << " " << ans[i][1] << "\n";
  return 0;
}
