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;
static const int INF = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
template <typename T, typename U>
static void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static void amax(T &x, U y) {
  if (x < y) x = y;
}
struct Sum {
  int min;
  long long sum;
  Sum() : min(INF), sum(0) {}
  explicit Sum(int w) : min(w), sum(w) {}
  Sum &operator+=(const Sum &that) {
    amin(min, that.min);
    sum += that.sum;
    return *this;
  }
};
int main() {
  int n;
  long long len;
  while (~scanf("%d%lld", &n, &len)) {
    vector<int> f(n);
    for (int i = 0; i < n; ++i) scanf("%d", &f[i]);
    vector<int> w(n);
    for (int i = 0; i < n; ++i) scanf("%d", &w[i]);
    vector<pair<int, Sum>> pow(n), npow(n);
    for (int(i) = 0; (i) < (int)(n); ++(i)) pow[i] = {f[i], Sum(w[i])};
    vector<pair<int, Sum>> ans(n);
    for (int(i) = 0; (i) < (int)(n); ++(i)) ans[i] = {i, Sum()};
    long long rem = len;
    for (long long p = 1; rem > 0; p *= 2) {
      if (rem & p) {
        for (int(i) = 0; (i) < (int)(n); ++(i)) {
          auto &x = pow[ans[i].first];
          ans[i].first = x.first;
          ans[i].second += x.second;
        }
        rem -= p;
      }
      npow = pow;
      for (int(i) = 0; (i) < (int)(n); ++(i)) {
        auto &x = pow[pow[i].first];
        npow[i].first = x.first;
        npow[i].second += x.second;
      }
      pow.swap(npow);
    }
    for (int(i) = 0; (i) < (int)(n); ++(i)) {
      const Sum &sum = ans[i].second;
      printf("%lld %d\n", sum.sum, sum.min);
    }
  }
  return 0;
}
