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;
long long int dp[35][100005];
long long int par[35][100005];
long long int mi[35][100005];
long long int a[100005];
long long int b[100005];
vector<long long int> bt;
int main() {
  long long int n, k;
  cin >> n >> k;
  for (long long int i = 0; i < n; i++) scanf("%lld", a + i);
  for (long long int i = 0; i < n; i++) scanf("%lld", b + i);
  for (long long int i = 0; i < n; i++) par[0][i] = a[i];
  for (long long int i = 1; i < 35; i++) {
    for (long long int j = 0; j < n; j++) par[i][j] = par[i - 1][par[i - 1][j]];
  }
  for (long long int i = 0; i < n; i++) dp[0][i] = b[i], mi[0][i] = b[i];
  for (long long int i = 1; i < 35; i++) {
    for (long long int j = 0; j < n; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i - 1][par[i - 1][j]];
      mi[i][j] = min(mi[i - 1][j], mi[i - 1][par[i - 1][j]]);
    }
  }
  for (long long int j = 0; j < 35; j++) {
    if (k & (1LL << j)) bt.push_back(j);
  }
  for (long long int i = 0; i < n; i++) {
    long long int ver = i;
    long long int dist = 0;
    long long int mii = 1e18 + 5;
    for (auto j : bt) {
      dist += dp[j][ver];
      mii = min(mii, mi[j][ver]);
      ver = par[j][ver];
    }
    printf("%lld %lld\n", dist, mii);
  }
  return 0;
}
