Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long t, n, i, j, k, x, y, z, ans, l, r, mid, m, s;
long long dp[100001][36], dp2[100001][36], dp3[100001][36];
void build() {
  for (i = 1; i <= 35; i++)
    for (j = 0; j < n; j++) dp[j][i] = dp[dp[j][i - 1]][i - 1];
  for (i = 1; i <= 35; i++)
    for (j = 0; j < n; j++)
      dp2[j][i] = min(dp2[dp[j][i - 1]][i - 1], dp2[j][i - 1]);
  for (i = 1; i <= 35; i++)
    for (j = 0; j < n; j++)
      dp3[j][i] = dp3[dp[j][i - 1]][i - 1] + dp3[j][i - 1];
}
int main() {
  ios::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    cin >> x;
    dp[i][0] = x;
  }
  for (i = 0; i < n; i++) {
    cin >> x;
    dp2[i][0] = dp3[i][0] = x;
  }
  build();
  for (i = 0; i < n; i++) {
    x = i;
    y = 1e18;
    z = 0;
    for (j = 35; j >= 0; j--)
      if (k & (1LL << j)) {
        y = min(dp2[x][j], y);
        z += dp3[x][j];
        x = dp[x][j];
      }
    cout << z << " " << y << "\n";
  }
}