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>
const long long maxn = 100100;
long long n, k;
long long fa[maxn][35][3];
long long lg[35], n2[35];
signed main() {
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i <= n; i++) {
    long long x;
    scanf("%lld", &x);
    fa[i][0][0] = x + 1;
  }
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &fa[i][0][1]);
    fa[i][0][2] = fa[i][0][1];
  }
  n2[0] = 1;
  for (long long i = 1; i < 35; i++) {
    lg[i] = lg[i - 1] + ((1 << lg[i - 1] == i));
    n2[i] = n2[i - 1] << 1;
  }
  for (long long j = 1; j < 35; j++) {
    for (long long i = 1; i <= n; i++) {
      fa[i][j][0] = fa[fa[i][j - 1][0]][j - 1][0];
      fa[i][j][1] = fa[fa[i][j - 1][0]][j - 1][1] + fa[i][j - 1][1];
      fa[i][j][2] = std::min(fa[i][j - 1][2], fa[fa[i][j - 1][0]][j - 1][2]);
    }
  }
  for (long long i = 1; i <= n; i++) {
    long long now = i, val = k, ans1 = 0, ans2 = 100000000000000000;
    for (long long j = 34; j >= 0; j--) {
      if (val - n2[j] >= 0) {
        val -= n2[j];
        ans1 += fa[now][j][1];
        ans2 = std::min(ans2, fa[now][j][2]);
        now = fa[now][j][0];
      }
    }
    printf("%lld %lld\n", ans1, ans2);
  }
}
