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 long long oo = 1e18 + 7, mod = 1e9 + 7;
const long long N = 1e5 + 5;
struct node {
  long long end, mx, tol;
} sparse[100005][34];
long long n, k, nxt[N], dist_loop[N], loop[N], lst[N], w[N];
bool vis[N];
signed main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> nxt[i];
    ++nxt[i];
  }
  for (long long i = 1; i <= n; i++) cin >> w[i];
  for (long long i = 1; i <= n; i++) {
    sparse[i][0].end = nxt[i];
    sparse[i][0].mx = w[i];
    sparse[i][0].tol = w[i];
  }
  for (long long i = 1; i <= 33; i++) {
    for (long long j = 1; j <= n; j++) {
      sparse[j][i].end = sparse[sparse[j][i - 1].end][i - 1].end;
      sparse[j][i].mx =
          min(sparse[j][i - 1].mx, sparse[sparse[j][i - 1].end][i - 1].mx);
      sparse[j][i].tol =
          sparse[j][i - 1].tol + sparse[sparse[j][i - 1].end][i - 1].tol;
    }
  }
  for (long long i = 1; i <= n; i++) {
    node ans;
    ans.end = i;
    ans.tol = 0;
    ans.mx = oo;
    for (long long j = 33; j >= 0; j--) {
      if (k & (1LL << j)) {
        ans.mx = min(ans.mx, sparse[ans.end][j].mx);
        ans.tol += sparse[ans.end][j].tol;
        ans.end = sparse[ans.end][j].end;
      }
    }
    cout << ans.tol << " " << ans.mx << "\n";
  }
}