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;
const int MAXN = 100050;
const long long mod = 1e9 + 7;
long long inp1[MAXN], inp2, inp3, ans = 1;
long long n, k, st[MAXN][40], sum[MAXN][40], mn[MAXN][40], mi[40];
int main() {
  mi[0] = 1;
  for (long long i = 1; i <= 34; i++) mi[i] = mi[i - 1] << 1;
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i <= n; i++) scanf("%lld", &inp1[i]);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &inp2);
    sum[i][0] = mn[i][0] = inp2;
    st[i][0] = inp1[i] + 1;
  }
  for (long long j = 1; j <= 34; j++) {
    for (long long i = 1; i <= n; i++) {
      st[i][j] = st[st[i][j - 1]][j - 1];
      sum[i][j] = sum[i][j - 1] + sum[st[i][j - 1]][j - 1];
      mn[i][j] = min(mn[i][j - 1], mn[st[i][j - 1]][j - 1]);
    }
  }
  for (long long i = 1; i <= n; i++) {
    long long now = i, tmp = k, ans1 = 0, ans2 = 1e12;
    while (tmp > 0) {
      long long pos = upper_bound(mi, mi + 35, tmp) - (mi + 1);
      ans1 += sum[now][pos];
      ans2 = min(ans2, mn[now][pos]);
      now = st[now][pos];
      tmp -= mi[pos];
    }
    printf("%lld %lld\n", ans1, ans2);
  }
  return 0;
}