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>
using namespace std;
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 | '0');
}
inline void wln(long long x) {
  write(x);
  puts("");
}
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
inline int Min(int x, int y) { return x > y ? y : x; }
long long to[300000], f[100005][60], v[300000], w[100005][60], sum[100005][59],
    ppp[59];
int main() {
  long long n = read(), k = read();
  ppp[0] = 1;
  for (int i = 1; i <= 46; ++i) ppp[i] = ppp[i - 1] * 2;
  for (long long i = 0; i < n; ++i) to[i] = read(), f[i][0] = to[i];
  for (long long i = 0; i < n; ++i)
    v[i] = read(), w[i][0] = v[i], sum[i][0] = v[i];
  for (long long i = 1; i <= 46; ++i)
    for (long long j = 0; j < n; ++j)
      f[j][i] = f[f[j][i - 1]][i - 1],
      w[j][i] = Min(w[j][i - 1], w[f[j][i - 1]][i - 1]),
      sum[j][i] = (long long)sum[j][i - 1] + sum[f[j][i - 1]][i - 1];
  for (long long i = 0; i < n; ++i) {
    long long now = k, p = 210000000000000000, x = i, q = 0;
    for (long long i = 46; i >= 0; --i)
      if (now >= ppp[i]) {
        now -= ppp[i];
        p = Min(p, w[x][i]);
        q += sum[x][i];
        x = f[x][i];
        if (now == 0) break;
      }
    printf("%lld %lld\n", q, p);
  }
  return 0;
}
