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 long long N = 1e5 + 100, P = 1e9 + 7, INF = 2e9 + 100;
const double ESP = 1. / 10000000000.;
struct aaa {
  long long p, val, minl;
  aaa() { p = val = minl = 0; }
};
aaa spr[36][N];
long long n, k, a;
void build() {
  for (int i = 1; i <= 35; i++) {
    for (int j = 1; j <= n; j++) {
      spr[i][j].p = spr[i - 1][spr[i - 1][j].p].p;
      spr[i][j].val = spr[i - 1][spr[i - 1][j].p].val + spr[i - 1][j].val;
      spr[i][j].minl =
          min(spr[i - 1][j].minl, spr[i - 1][spr[i - 1][j].p].minl);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    spr[0][i].p = a + 1;
  }
  for (int i = 1; i <= n; i++) {
    cin >> a;
    spr[0][i].val = a;
    spr[0][i].minl = a;
  }
  build();
  for (int i = 1; i <= n; i++) {
    long long kk = k;
    long long v = i;
    long long minl = INF;
    long long sum = 0;
    for (int j = 35; j >= 0; j--) {
      if (kk - ((long long)1 << (long long)j) >= 0) {
        minl = min(minl, spr[j][v].minl);
        sum += spr[j][v].val;
        v = spr[j][v].p;
        kk -= ((long long)1 << (long long)j);
      }
    }
    cout << sum << ' ' << minl << endl;
  }
}
