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;
const int N = 100005 + 10;
long long fa[N][36];
long long son[N][36];
long long minn[N][36];
long long n, k;
void sovle(long long u) {
  long long ans = 0, Min = 1e18;
  long long K = k;
  for (long long i = 35; i >= 0; i--) {
    if ((1ll << i) <= K) {
      ans += fa[u][i];
      Min = min(Min, minn[u][i]);
      u = son[u][i];
      K -= (1ll << i);
    }
  }
  printf("%lld %lld\n", ans, Min);
}
signed main() {
  scanf("%lld%lld", &n, &k);
  for (long long i = 0; i < n; i++) scanf("%lld", &son[i][0]);
  for (long long i = 0; i < n; i++) {
    scanf("%lld", &fa[i][0]);
    minn[i][0] = fa[i][0];
  }
  for (long long j = 1; j <= 35; j++) {
    for (long long i = 0; i < n; i++) {
      son[i][j] = son[son[i][j - 1]][j - 1];
      fa[i][j] = fa[i][j - 1] + fa[son[i][j - 1]][j - 1];
      minn[i][j] = min(minn[i][j - 1], minn[son[i][j - 1]][j - 1]);
    }
  }
  for (long long i = 0; i < n; i++) {
    sovle(i);
  }
  return 0;
}