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;
using ll = long long;
using P = pair<int, int>;
using vi = vector<int>;
using vc = vector<char>;
using vb = vector<bool>;
using vs = vector<string>;
using vll = vector<long long>;
using vp = vector<pair<int, int>>;
using vvi = vector<vector<int>>;
using vvc = vector<vector<char>>;
using vvll = vector<vector<long long>>;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
const int INF = 1001001001;
int to[40][100005], mn[40][100005];
ll cost[40][100005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  ll k;
  cin >> n >> k;
  for (int i = 0; i < (int)(n); i++) {
    cin >> to[0][i];
  }
  for (int i = 0; i < (int)(n); i++) {
    cin >> cost[0][i];
    mn[0][i] = cost[0][i];
  }
  for (int t = 0; t < (int)(39); t++) {
    for (int i = 0; i < (int)(n); i++) {
      to[t + 1][i] = to[t][to[t][i]];
    }
  }
  for (int t = 0; t < (int)(39); t++) {
    for (int i = 0; i < (int)(n); i++) {
      cost[t + 1][i] = cost[t][i] + cost[t][to[t][i]];
      mn[t + 1][i] = min(mn[t][i], mn[t][to[t][i]]);
    }
  }
  vll s(n);
  vi m(n, INF);
  for (int i = 0; i < (int)(n); i++) {
    int now = i;
    for (int j = 0; j < (int)(40); j++) {
      if (k >> j & 1) {
        s[i] += cost[j][now];
        chmin(m[i], mn[j][now]);
        now = to[j][now];
      }
    }
  }
  for (int i = 0; i < (int)(n); i++) cout << s[i] << " " << m[i] << endl;
}
