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>
const int Inf = 2 * 1000 * 1000 * 1000;
long long LINF = (long long)4e18;
using namespace std;
const int nax = 1000 * 100 + 5;
const int LOGN = 60;
long long nt[LOGN][nax];
long long s[LOGN][nax];
long long mn[LOGN][nax];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  vector<long long> vals(n);
  for (auto &x : a) cin >> x;
  for (auto &x : vals) cin >> x;
  for (int i = 0; i < n; i++) {
    nt[0][i] = a[i];
    s[0][i] = vals[i];
    mn[0][i] = vals[i];
  }
  for (int i = 1; i < LOGN; i++) {
    for (int j = 0; j < n; j++) {
      nt[i][j] = nt[i - 1][nt[i - 1][j]];
      s[i][j] = s[i - 1][j] + s[i - 1][nt[i - 1][j]];
      mn[i][j] = min(mn[i - 1][j], mn[i - 1][nt[i - 1][j]]);
    }
  }
  for (int i = 0; i < n; i++) {
    long long minval = 4e18;
    long long sum = 0;
    int v = i;
    for (int j = 0; j < LOGN; j++) {
      if (k & (1LL << j)) {
        sum += s[j][v];
        minval = min(minval, mn[j][v]);
        v = nt[j][v];
      }
    }
    cout << sum << ' ' << minval << '\n';
  }
  return 0;
}
