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 M = 100005;
int n, m, x, y, f[M], we[M];
long long k, dp[40][M], w[40][M], mn[40][M];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> f[i];
    dp[0][i] = f[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> we[i];
    w[0][i] = we[i];
    mn[0][i] = we[i];
  }
  for (int k = 1; k < 40; k++)
    for (int u = 0; u < n; u++) {
      dp[k][u] = dp[k - 1][dp[k - 1][u]];
      w[k][u] = w[k - 1][u] + w[k - 1][dp[k - 1][u]];
      mn[k][u] = min(mn[k - 1][u], mn[k - 1][dp[k - 1][u]]);
    }
  for (int i = 0; i < n; i++) {
    long long sum = 0, mnn = 2e18, t = i;
    for (int j = 0; j < 40; j++) {
      if (k & (1LL << j)) {
        sum += w[j][t];
        mnn = min(mnn, mn[j][t]);
        t = dp[j][t];
      }
    }
    cout << sum << " " << mnn << endl;
  }
  return 0;
}
