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 = 100005;
const int K = 40;
int n;
long long k;
int nx[K][N];
long long sum[K][N];
int mn[K][N];
int a[N];
int go[50], len = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> nx[0][i];
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) sum[0][i] = mn[0][i] = a[i];
  for (int step = 1; step <= 35; ++step) {
    for (int i = 0; i < n; ++i) {
      nx[step][i] = nx[step - 1][nx[step - 1][i]];
      sum[step][i] = sum[step - 1][nx[step - 1][i]] + sum[step - 1][i];
      mn[step][i] = min(mn[step - 1][nx[step - 1][i]], mn[step - 1][i]);
    }
  }
  for (int jump = 35; jump >= 0; --jump) {
    if ((1ll << jump) <= k) {
      k -= (1ll << jump);
      go[++len] = jump;
    }
  }
  for (int i = 0; i < n; ++i) {
    long long s = 0;
    int m = 1000000000;
    int v = i;
    for (int it = 1; it <= len; ++it) {
      s += sum[go[it]][v];
      m = min(m, mn[go[it]][v]);
      v = nx[go[it]][v];
    }
    cout << s << " " << m << endl;
  }
  return 0;
}