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 int MOD = 1000000007;
template <class T, class U>
inline T tmin(T a, U b) {
  return (a < b) ? a : b;
}
template <class T, class U>
inline T tmax(T a, U b) {
  return (a > b) ? a : b;
}
template <class T, class U>
inline void amax(T &a, U b) {
  if (b > a) a = b;
}
template <class T, class U>
inline void amin(T &a, U b) {
  if (b < a) a = b;
}
template <class T>
inline T tabs(T a) {
  return (a > 0) ? a : -a;
}
template <class T>
T gcd(T a, T b) {
  while (b != 0) {
    T c = a;
    a = b;
    b = c % b;
  }
  return a;
}
const int N = 100005, D = 35;
int f[N][D], m[N][D];
long long s[N][D];
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  int n;
  long long k;
  cin >> n >> k;
  for (int i = (0); i < (n); ++i) cin >> f[i][0];
  for (int i = (0); i < (n); ++i) {
    cin >> m[i][0];
    s[i][0] = m[i][0];
  }
  for (int d = (1); d < (D); ++d) {
    for (int i = (0); i < (n); ++i) {
      f[i][d] = f[f[i][d - 1]][d - 1];
      s[i][d] = s[i][d - 1] + s[f[i][d - 1]][d - 1];
      m[i][d] = tmin(m[i][d - 1], m[f[i][d - 1]][d - 1]);
    }
  }
  for (int i = (0); i < (n); ++i) {
    long long tot = 0, rem = k;
    int mini = INT_MAX, cur = i;
    for (int d = (D - 1); d > (-1); --d)
      if (rem >= (1LL << d)) {
        amin(mini, m[cur][d]);
        tot += s[cur][d];
        cur = f[cur][d];
        rem -= 1LL << d;
      }
    printf("%lld %d\n", tot, mini);
  }
  return 0;
}
