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 = 100010, maxlog = 36;
int n;
long long k;
struct f {
  long long sum;
  int to, mn;
} dp[maxlog][maxn];
int main() {
  scanf("%d %lld", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &dp[0][i].to);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &dp[0][i].mn);
    dp[0][i].sum = dp[0][i].mn;
  }
  for (int i = 1; i < maxlog; i++) {
    for (int j = 0; j < n; j++) {
      dp[i][j].to = dp[i - 1][dp[i - 1][j].to].to;
      dp[i][j].sum = dp[i - 1][j].sum + dp[i - 1][dp[i - 1][j].to].sum;
      dp[i][j].mn = min(dp[i - 1][j].mn, dp[i - 1][dp[i - 1][j].to].mn);
    }
  }
  for (int j = 0; j < n; j++) {
    f r = {0ll, j, 1000000000};
    for (int i = 0; i < maxlog; i++) {
      if (k & (1ll << i)) {
        r.sum += dp[i][r.to].sum;
        r.mn = min(r.mn, dp[i][r.to].mn);
        r.to = dp[i][r.to].to;
      }
    }
    printf("%lld %d\n", r.sum, r.mn);
  }
  return 0;
}