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;
const int N = 100005;
int out[N], w[N], dp[N][40];
long long sum[N][40], mn[N][40];
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> out[i];
    dp[i][0] = out[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> w[i];
    sum[i][0] = w[i];
    mn[i][0] = w[i];
  }
  for (int k = 1; k < 40; ++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 si = 0;
    int cur = i;
    long long rem = k;
    long long mi = 1e18;
    for (long long p = 39; p >= 0; --p) {
      if ((1ll << p) <= rem) {
        rem -= (1ll << p);
        mi = min(mi, mn[cur][p]);
        si += sum[cur][p];
        cur = dp[cur][p];
      }
    }
    cout << si << ' ' << mi << '\n';
  }
  return 0;
}
