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 inf = 1e9;
const long long inf64 = 1e18;
const long long MOD = inf + 7;
const long long N = 1e5 + 5;
const long long LG = 35;
long long mn[N][LG];
long long sum[N][LG];
long long par[N][LG];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 0; i < N; i++)
    for (long long j = 0; j < LG; j++) mn[i][j] = LLONG_MAX;
  long long n, k;
  cin >> n >> k;
  long long p[n], w[n];
  for (long long i = 0; i < n; i++) {
    cin >> p[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> w[i];
    mn[i][0] = w[i];
    sum[i][0] = w[i];
    par[i][0] = p[i];
  }
  for (long long i = 1; i < LG; i++) {
    for (long long j = 0; j < n; j++) {
      par[j][i] = par[par[j][i - 1]][i - 1];
      sum[j][i] += sum[par[j][i - 1]][i - 1] + sum[j][i - 1];
      mn[j][i] = min(mn[par[j][i - 1]][i - 1], mn[j][i - 1]);
    }
  }
  for (long long i = 0; i < n; i++) {
    long long res = 0, mini = LLONG_MAX;
    long long node = i;
    for (long long j = 0; j < LG; j++) {
      if (k >> j & 1) {
        res += sum[node][j];
        mini = min(mini, mn[node][j]);
        node = par[node][j];
      }
    }
    cout << res << " " << mini << "\n";
  }
}
