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 ii = pair<ll, ll>;
const int MAX{100010}, MAXK{50};
const ll MOD{1000000007}, oo{1LL << 62};
ll ps[MAX][MAXK], ms[MAX][MAXK], s[MAX][MAXK];
vector<ii> solve(int N, ll K, const vector<ll>& fs, const vector<ll>& cs) {
  for (int i = 0; i < N; ++i) {
    ps[i][1] = fs[i];
    ms[i][1] = s[i][1] = cs[i];
  }
  for (int k = 2; k < MAXK; ++k) {
    for (int i = 0; i < N; ++i) {
      auto j = ps[i][k - 1];
      ps[i][k] = ps[j][k - 1];
      ms[i][k] = min(ms[i][k - 1], ms[j][k - 1]);
      s[i][k] = s[i][k - 1] + s[j][k - 1];
    }
  }
  vector<ii> ans(N, ii(0, oo));
  vector<int> ns(N);
  iota(ns.begin(), ns.end(), 0);
  for (int k = 1; k < MAXK; ++k) {
    ll mask = 1LL << (k - 1);
    if (mask & K) {
      for (int i = 0; i < N; ++i) {
        auto j = ns[i];
        ans[i].first += s[j][k];
        ans[i].second = min(ans[i].second, ms[j][k]);
        ns[i] = ps[j][k];
      }
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  ll N, K;
  cin >> N >> K;
  vector<ll> fs(N), cs(N);
  for (int i = 0; i < N; ++i) cin >> fs[i];
  for (int i = 0; i < N; ++i) cin >> cs[i];
  auto ans = solve(N, K, fs, cs);
  for (auto [s, m] : ans) cout << s << ' ' << m << '\n';
  return 0;
}
