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>
inline long long read() {
  long long x = 0;
  char c = getchar();
  bool k = false;
  for (; !isdigit(c); c = getchar()) k |= c == '-';
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  if (k) x = -x;
  return x;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const long long maxn = 1e5 + 10;
long long sum[maxn][40];
long long min[maxn][40];
long long fa[maxn][40];
long long n, k;
long long num[40], pos;
signed main() {
  n = read();
  k = read();
  for (long long i = 1; i <= n; i++) fa[i][0] = read() + 1;
  for (long long i = 1; i <= n; i++) sum[i][0] = min[i][0] = read();
  for (long long i = 1; i <= 39; i++)
    for (long long j = 1; j <= n; j++)
      fa[j][i] = fa[fa[j][i - 1]][i - 1],
      sum[j][i] = sum[j][i - 1] + sum[fa[j][i - 1]][i - 1],
      min[j][i] = std::min(min[j][i - 1], min[fa[j][i - 1]][i - 1]);
  for (long long i = 0; i <= 39; i++)
    if (k & (long long)std::pow(2, i)) num[++pos] = i;
  for (long long i = 1; i <= n; i++) {
    long long p = i;
    long long ans1 = 0, ans2 = 0x3f3f3f3f;
    for (long long j = 1; j <= pos; j++) {
      ans1 += sum[p][num[j]];
      ans2 = std::min(ans2, min[p][num[j]]);
      p = fa[p][num[j]];
    }
    write(ans1);
    putchar(' ');
    write(ans2);
    putchar('\n');
  }
  return 0;
}
