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;
int read() {
  char ch = getchar();
  int x = 0;
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x;
}
const int maxn = 1e5 + 5, maxl = 40, inf = 1e9;
int N, x, mn, f[maxn][maxl], v[maxn][maxl];
long long w[maxn][maxl], K, sm;
int main() {
  scanf("%d%lld", &N, &K);
  for (int i = 1; i <= N; ++i) f[i][0] = read() + 1;
  for (int i = 1; i <= N; ++i) w[i][0] = v[i][0] = read();
  for (int i = 1; i < maxl; ++i)
    for (int j = 1; j <= N; ++j) {
      f[j][i] = f[f[j][i - 1]][i - 1];
      w[j][i] = w[f[j][i - 1]][i - 1] + w[j][i - 1];
      v[j][i] = min(v[f[j][i - 1]][i - 1], v[j][i - 1]);
    }
  for (int i = 1; i <= N; ++i) {
    x = i, mn = inf, sm = 0;
    for (int j = 0; j < maxl; ++j)
      if (K & (1LL << j)) {
        sm += w[x][j];
        mn = min(mn, v[x][j]);
        x = f[x][j];
      }
    printf("%lld %d\n", sm, mn);
  }
  return 0;
}