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 = 100100;
const int MAXLOGK = 36;
int N;
long long K;
int F[MAXN][MAXLOGK];
long long W[MAXN][MAXLOGK];
long long sum[MAXN][MAXLOGK];
int cur[MAXN];
long long ans[MAXN][2];
int main() {
  ios::sync_with_stdio(0);
  cin >> N >> K;
  for (int i = 0; i < N; i++) cin >> F[i][0];
  for (int i = 0; i < N; i++) {
    cin >> W[i][0];
    sum[i][0] = W[i][0];
  }
  for (int j = 1; j < MAXLOGK; j++)
    for (int i = 0; i < N; i++) {
      F[i][j] = F[F[i][j - 1]][j - 1];
      W[i][j] = min(W[i][j - 1], W[F[i][j - 1]][j - 1]);
      sum[i][j] = sum[i][j - 1] + sum[F[i][j - 1]][j - 1];
    }
  for (int i = 0; i < N; i++) cur[i] = i, ans[i][1] = 1 << 30;
  for (int j = MAXLOGK - 1; j >= 0; j--)
    if ((K & (1LL << j)))
      for (int i = 0; i < N; i++) {
        ans[i][0] += sum[cur[i]][j];
        ans[i][1] = min(ans[i][1], W[cur[i]][j]);
        cur[i] = F[cur[i]][j];
      }
  for (int i = 0; i < N; i++) cout << ans[i][0] << " " << ans[i][1] << "\n";
  return 0;
}