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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx2,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vi>;
using pi = pair<ll, ll>;
const int mod = 1e9 + 7;
const ll inf = 1e18;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll n, k, mn[40][100100], sm[40][100100], p[40][100100];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> p[0][i];
  for (int i = 0; i < n; i++) cin >> mn[0][i], sm[0][i] = mn[0][i];
  for (int i = 1; i < 34; i++) {
    for (int j = 0; j < n; j++) {
      p[i][j] = p[i - 1][p[i - 1][j]];
      sm[i][j] = sm[i - 1][j] + sm[i - 1][p[i - 1][j]];
      mn[i][j] = min(mn[i - 1][j], mn[i - 1][p[i - 1][j]]);
    }
  }
  for (int i = 0; i < n; i++) {
    ll u = i, m = LLONG_MAX, s = 0;
    for (int j = 34; j--;) {
      if (k & (1ll << j)) {
        m = min(m, mn[j][u]);
        s = s + sm[j][u];
        u = p[j][u];
      }
    }
    cout << s << " " << m << "\n";
  }
}
