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;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long int ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
template <class T>
inline T bpow(T p, T e) {
  long long int ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p);
    p = (p * p);
  }
  return (T)ret;
}
int toInt(string s) {
  int sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
int toLlint(string s) {
  long long int sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
const double eps = 1e-9;
int ts, kk = 1;
int n, mx;
long long int p[3][40][100005], m, q;
int main() {
  int i, j, k, t;
  memset(p, -1, sizeof(p));
  scanf("%d%lld", &n, &m);
  for (i = 0; i < n; i++) {
    scanf("%d", &k);
    p[0][0][i] = k;
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &k);
    p[1][0][i] = k;
    p[2][0][i] = k;
  }
  for (i = 1; (1LL << i) <= m; i++) {
    for (j = 0; j < n; j++)
      if (p[0][i - 1][j] != -1) {
        k = p[0][i - 1][j];
        p[0][i][j] = p[0][i - 1][k];
        p[1][i][j] = p[1][i - 1][j] + p[1][i - 1][k];
        p[2][i][j] = min(p[2][i - 1][j], p[2][i - 1][k]);
      }
  }
  mx = i - 1;
  for (j = 0; j < n; j++) {
    q = m;
    long long int r1, r2;
    r1 = 0;
    r2 = (LLONG_MAX / 10LL);
    k = j;
    for (i = mx; i > -1; i--) {
      if ((1LL << i) <= q) {
        q -= (1LL << i);
        r1 += p[1][i][k];
        r2 = min(r2, p[2][i][k]);
        k = p[0][i][k];
      }
    }
    printf("%lld %lld\n", r1, r2);
  }
  return 0;
}
