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;
const long long N = 1e5 + 5, M = 47;
long long g[N][M], f[N][M], a[N][M], n, k;
signed main() {
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i][0]);
    a[i][0]++;
  }
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &f[i][0]);
    g[i][0] = f[i][0];
  }
  for (long long j = 1; j <= 45; j++) {
    for (long long i = 1; i <= n; i++) {
      long long s = a[i][j - 1];
      a[i][j] = a[s][j - 1];
      f[i][j] = f[i][j - 1] + f[s][j - 1];
      g[i][j] = min(g[i][j - 1], g[s][j - 1]);
    }
  }
  for (long long i = 1; i <= n; i++) {
    long long res = 0, minx = 1ll << 45, now = i;
    for (long long j = 45; j >= 0; j--) {
      if ((k >> j) & 1LL) {
        res += f[now][j];
        minx = min(minx, g[now][j]);
        now = a[now][j];
      }
    }
    printf("%lld %lld\n", res, minx);
  }
  return 0;
}
