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;
const double eps = 1e-9;
const int MAXN = 100100;
const int MAXP = 35;
int to[MAXP][MAXN];
int mini[MAXP][MAXN];
long long sum[MAXP][MAXN];
int N;
long long K;
int main() {
  ios::sync_with_stdio(false);
  cin >> N >> K;
  for (int i = (0); i < (N); i++) cin >> to[0][i];
  for (int i = (0); i < (N); i++) {
    cin >> mini[0][i];
    sum[0][i] = mini[0][i];
  }
  int P = 0;
  while ((1LL << P) < K) {
    P++;
    for (int i = (0); i < (N); i++) {
      to[P][i] = to[P - 1][to[P - 1][i]];
      sum[P][i] = sum[P - 1][i] + sum[P - 1][to[P - 1][i]];
      mini[P][i] = min(mini[P - 1][i], mini[P - 1][to[P - 1][i]]);
    }
    assert(P < MAXP);
  }
  for (int i = (0); i < (N); i++) {
    int c = i;
    long long csum = 0;
    int cmin = 1e9;
    for (int p = (0); p < (P + 1); p++)
      if ((K >> p) & 1) {
        csum += sum[p][c];
        cmin = min(cmin, mini[p][c]);
        c = to[p][c];
      }
    cout << csum << " " << cmin << endl;
  }
  return 0;
}
