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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (long long)(ch - '0');
    ch = getchar();
  }
  return x * f;
}
long long f[100009][37], minn[100009][37], sum[100009][37];
long long n, m, k, cnt, ans;
int main() {
  n = read(), k = read();
  for (int i = 0; i < n; i++) f[i][0] = read();
  for (int i = 0; i < n; i++) minn[i][0] = read(), sum[i][0] = minn[i][0];
  for (int j = 1; j <= 36; j++)
    for (int i = 0; i < n; i++) {
      f[i][j] = f[f[i][j - 1]][j - 1];
      sum[i][j] = sum[f[i][j - 1]][j - 1] + sum[i][j - 1];
      minn[i][j] = min(minn[i][j - 1], minn[f[i][j - 1]][j - 1]);
    }
  for (int i = 0; i < n; i++) {
    long long now = i, mn = 0x3f3f3f3f;
    ans = 0;
    for (int j = 36; j >= 0; j--) {
      if ((1LL << j) & k) {
        ans += sum[now][j];
        mn = min(mn, minn[now][j]);
        now = f[now][j];
      }
    }
    printf(i == n - 1 ? "%lld %lld" : "%lld %lld\n", ans, mn);
  }
  return 0;
}