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 int MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
long long mini[35][MAXN];
long long sum[35][MAXN];
int P[35][MAXN];
long long POWER[50];
pair<long long, long long> go(int idx, long long k) {
  if (!k) return make_pair(0, MOD);
  int p = -1;
  for (int i = 0; POWER[i] <= k and p == -1; i++)
    if (k & POWER[i]) p = i;
  pair<long long, long long> ret = {0, MOD};
  ret.first += sum[p][idx];
  ret.second = min(ret.second, mini[p][idx]);
  pair<long long, long long> g = go(P[p][idx], k - POWER[p]);
  ret.first += g.first;
  ret.second = min(ret.second, g.second);
  return ret;
}
int main() {
  POWER[0] = 1;
  for (int i = 1; i < 50; i++) POWER[i] = POWER[i - 1] << 1LL;
  int n;
  cin >> n;
  long long k;
  cin >> k;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    P[0][i] = x;
  }
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    mini[0][i] = x;
    sum[0][i] = x;
  }
  for (int i = 1; POWER[i] <= k; i++) {
    for (int j = 0; j < n; j++) {
      mini[i][j] = min(mini[i - 1][j], mini[i - 1][P[i - 1][j]]);
      sum[i][j] = sum[i - 1][j] + sum[i - 1][P[i - 1][j]];
      P[i][j] = P[i - 1][P[i - 1][j]];
    }
  }
  for (int i = 0; i < n; i++) {
    pair<long long, long long> ans = go(i, k);
    printf("%lld %lld\n", ans.first, ans.second);
  }
  return 0;
}
