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;
template <typename T>
inline void read(T &FF) {
  int RR = 1;
  FF = 0;
  char CH = getchar();
  for (; !isdigit(CH); CH = getchar())
    if (CH == '-') RR = -RR;
  for (; isdigit(CH); CH = getchar()) FF = FF * 10 + CH - 48;
  FF *= RR;
}
inline void file(string str) {
  freopen((str + ".in").c_str(), "r", stdin);
  freopen((str + ".out").c_str(), "w", stdout);
}
const int N = 1e5 + 10, Log = 40;
int son[N][Log + 1], n;
long long k, val[N][Log + 1], sb[N][Log + 1];
void solve(int x) {
  long long pi = 1, ri = 0, minx = INT_MAX, res = 0;
  for (long long i = 1; i <= 40; i++) pi *= 2;
  for (int i = Log; i >= 0; i--, pi /= 2)
    if (ri + pi <= k) {
      minx = min(minx, sb[x][i]);
      res += val[x][i];
      x = son[x][i];
      ri += pi;
    }
  printf("%lld %lld\n", res, minx);
}
int main() {
  read(n), read(k);
  for (int i = 1; i <= n; i++) read(son[i][0]), son[i][0] += 1;
  for (int i = 1; i <= n; i++) read(val[i][0]), sb[i][0] = val[i][0];
  for (int i = 1; i <= Log; i++)
    for (int j = 1; j <= n; j++) {
      son[j][i] = son[son[j][i - 1]][i - 1];
      sb[j][i] = min(sb[j][i - 1], sb[son[j][i - 1]][i - 1]);
      val[j][i] = val[j][i - 1] + val[son[j][i - 1]][i - 1];
    }
  for (int i = 1; i <= n; i++) solve(i);
  return 0;
}