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;
int n, f[100010], w[100010], c;
long long k;
int dp1[36][200010];
long long dp2[36][200010];
int dp3[36][200010];
void init() {
  for (int i = 1; i < 35; i++) {
    for (int j = 0; j < n; j++) {
      dp1[i][j] = dp1[i - 1][dp1[i - 1][j]];
      dp2[i][j] = dp2[i - 1][j] + dp2[i - 1][dp1[i - 1][j]];
      dp3[i][j] = min(dp3[i - 1][j], dp3[i - 1][dp1[i - 1][j]]);
    }
  }
}
bool v[200010];
long long W[200010];
long long M[200010];
long long binaryLift(int cur, long long& mini) {
  long long logi = 0;
  long long K = k;
  for (; (1LL << logi) <= K; logi++)
    ;
  logi--;
  long long tw = 0LL;
  for (long long i = logi; i >= 0; i--) {
    if (K - (1LL << i) >= 0LL) {
      K -= (1LL << i);
      tw += dp2[i][cur];
      mini = min(mini, (long long)dp3[i][cur]);
      cur = dp1[i][cur];
    }
  }
  return tw;
}
int main() {
  scanf("%d %I64d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &f[i]);
  for (int i = 0; i < n; i++) scanf("%d", &w[i]);
  for (int i = 0; i < n; i++) {
    int next = f[i];
    dp1[0][i] = next;
    dp2[0][i] = w[i];
    dp3[0][i] = w[i];
  }
  init();
  for (int i = 0; i < n; i++) {
    M[i] = w[i];
    W[i] = binaryLift(i, M[i]);
  }
  for (int i = 0; i < n; i++) printf("%I64d %I64d\n", W[i], M[i]);
  return 0;
}