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;
long long up[100005][40][3];
int main() {
  long long(n), (k);
  scanf("%lld%lld", &(n), &(k));
  ;
  for (long long i = 0; i < (100005); i++)
    for (long long j = 0; j < (40); j++)
      for (long long k = 0; k < (3); k++) up[i][j][k] = 0;
  vector<pair<long long, long long> > par(n);
  for (long long i = 0; i < (n); i++) {
    long long(a);
    scanf("%lld", &(a));
    ;
    par[i].first = a;
  }
  for (long long i = 0; i < (n); i++) {
    long long(a);
    scanf("%lld", &(a));
    ;
    par[i].second = a;
  }
  for (long long i = 0; i < (n); i++) {
    up[i][0][0] = par[i].first;
    up[i][0][1] = par[i].second;
    up[i][0][2] = par[i].second;
  }
  for (long long j = 1; j < (40); j++) {
    for (long long i = 0; i < (n); i++) {
      up[i][j][0] = up[up[i][j - 1][0]][j - 1][0];
      up[i][j][1] = min(up[up[i][j - 1][0]][j - 1][1], up[i][j - 1][1]);
      up[i][j][2] = up[up[i][j - 1][0]][j - 1][2] + up[i][j - 1][2];
    }
  }
  long long c = 1;
  for (long long i = 0; i < (n); i++) {
    long long ans1 = 0, ans2 = (long long)1e18;
    long long rem = k, t = i;
    for (long long j = 39; j >= (0); j--) {
      if ((c << j) <= rem) {
        ans1 += up[t][j][2];
        ans2 = min(ans2, up[t][j][1]);
        t = up[t][j][0];
        rem -= (c << j);
      }
    }
    printf("%lld %lld\n", ans1, ans2);
  }
  return 0;
}