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 N = 100010, LOGN = 50;
int dp[LOGN][N];
long long dp2[LOGN][N];
long long dp3[LOGN][N];
int n;
long long k;
void solve(int u) {
  long long tmp = k;
  long long ans = 0;
  long long ans2 = 1e18;
  for (int i = LOGN; i >= 0; i--) {
    if (((long long)1 << (long long)i) <= tmp) {
      tmp -= ((long long)1 << (long long)i);
      ans += dp2[i][u];
      ans2 = min(ans2, dp3[i][u]);
      u = dp[i][u];
    }
  }
  printf("%I64d %I64d\n", ans, ans2);
}
int main() {
  scanf("%d%I64d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &dp[0][i]);
  }
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    dp2[0][i] = a;
    dp3[0][i] = a;
  }
  for (int j = 1; ((long long)1 << (long long)j) <= k; j++) {
    for (int i = 0; i < n; i++) {
      dp[j][i] = dp[j - 1][dp[j - 1][i]];
      dp2[j][i] = dp2[j - 1][dp[j - 1][i]] + dp2[j - 1][i];
      dp3[j][i] = min(dp3[j - 1][dp[j - 1][i]], dp3[j - 1][i]);
    }
  }
  for (int i = 0; i < n; i++) {
    solve(i);
  }
  return 0;
}