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;
const int N = 1e5 + 5, mod = 1e9 + 7;
int n, v[N], w[N];
long long d;
struct node {
  int c, mn;
  long long sm;
  node() {
    c = 0, mn = 1e9;
    sm = 0;
  }
} dp[N][60];
int main() {
  scanf("%d%lld", &n, &d);
  for (int i = 0; i < n; i++) scanf("%d", &v[i]);
  for (int i = 0; i < n; i++) scanf("%d", &w[i]);
  for (int i = 0; i < n; i++)
    dp[i][0].c = v[i], dp[i][0].mn = dp[i][0].sm = w[i];
  for (int k = 1; k < 50; k++)
    for (int i = 0; i < n; i++) {
      dp[i][k].c = dp[dp[i][k - 1].c][k - 1].c;
      dp[i][k].mn = min(dp[dp[i][k - 1].c][k - 1].mn, dp[i][k - 1].mn);
      dp[i][k].sm = dp[dp[i][k - 1].c][k - 1].sm + dp[i][k - 1].sm;
    }
  for (int i = 0; i < n; i++) {
    int id = i, mn = 1e9;
    long long sm = 0;
    for (int k = 49; k >= 0; k--) {
      if (!(d & (1LL << k))) continue;
      sm += dp[id][k].sm;
      mn = min(mn, dp[id][k].mn);
      id = dp[id][k].c;
    }
    cout << sm << " " << mn << endl;
  }
}
