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;
struct node {
  int to;
  long long sumW;
  int minW;
};
vector<node> mul(const vector<node>& b, const vector<node>& a) {
  int n = a.size();
  vector<node> ans(n);
  for (int i = 0; i < n; ++i) {
    ans[i].to = b[a[i].to].to;
    ans[i].sumW = a[i].sumW + b[a[i].to].sumW;
    ans[i].minW = min(a[i].minW, b[a[i].to].minW);
  }
  return ans;
}
int main() {
  int n;
  long long k;
  cin >> n >> k;
  vector<node> f(n);
  for (int i = 0; i < n; ++i) cin >> f[i].to;
  for (int i = 0; i < n; ++i) {
    int w;
    cin >> w;
    f[i].sumW = f[i].minW = w;
  }
  vector<node> ans = f;
  k--;
  while (k) {
    if (k & 1) ans = mul(ans, f);
    f = mul(f, f);
    k >>= 1;
  }
  for (int i = 0; i < n; ++i) cout << ans[i].sumW << " " << ans[i].minW << endl;
  return 0;
}