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;
typedef struct node {
  int p;
  long long int w;
  int m;
  node(){};
} node;
int main() {
  int N;
  long long int K;
  cin >> N >> K;
  int p[N];
  int w[N];
  for (int i = 0; i < N; i++) scanf("%d", p + i);
  for (int j = 0; j < N; j++) scanf("%d", w + j);
  node dp[35][N];
  for (int i = 0; i < 35; i++) {
    for (int j = 0; j < N; j++) {
      if (i == 0) {
        dp[i][j].p = p[j];
        dp[i][j].w = w[j];
        dp[i][j].m = w[j];
      } else {
        dp[i][j].p = dp[i - 1][dp[i - 1][j].p].p;
        dp[i][j].w = dp[i - 1][j].w + dp[i - 1][dp[i - 1][j].p].w;
        dp[i][j].m = min(dp[i - 1][j].m, dp[i - 1][dp[i - 1][j].p].m);
      }
    }
  }
  for (int i = 0; i < N; i++) {
    long long int wt = 0, minm = 2e18;
    int x = i;
    for (int j = 0; j < 35; j++) {
      long long int bit = (long long)1 << j;
      if (K & bit) {
        wt += dp[j][x].w;
        minm = min(minm, (long long)dp[j][x].m);
        x = dp[j][x].p;
      }
    }
    cout << wt << " " << minm << endl;
  }
  return 0;
}
