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>
template <typename A, typename B>
inline void umin(A &x, B y) {
  if (x > y) x = y;
}
template <typename A, typename B>
inline void umax(A &x, B y) {
  if (x < y) x = y;
}
const int N = 200005;
using namespace std;
long long k;
int n;
long long sum[N][35];
int f[N], w[N];
int to[N][35], val[N][35];
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> to[i][0];
  for (int i = 0; i < n; i++) {
    cin >> val[i][0];
    sum[i][0] = val[i][0];
  }
  for (int j = 0; j < 33; j++)
    for (int i = 0; i < n; i++) {
      to[i][j + 1] = to[to[i][j]][j];
      sum[i][j + 1] = sum[i][j] + sum[to[i][j]][j];
      val[i][j + 1] = min(val[i][j], val[to[i][j]][j]);
    }
  for (int i = 0; i < n; i++) {
    int x = i;
    long long s = 0, m = 1000000000;
    for (int j = 33; j >= 0; j--) {
      if (k & (1LL << j)) {
        s += sum[x][j];
        umin(m, val[x][j]);
        x = to[x][j];
      }
    }
    cout << s << " " << m << "\n";
  }
}
