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 int mod = 1000000000 + 7;
const long long int N = 100000 + 6;
const long long int inf = 1e18 + 1e17;
long long int powm(long long int a, long long int b) {
  a = a % mod;
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return (res % mod);
}
long long int modInverse(long long int a, long long int m) {
  long long int m0 = m;
  long long int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long int q = a / m;
    long long int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long int p[37][N];
long long int sum[37][N];
long long int low[37][N];
void solve() {
  long long int n, k;
  cin >> n >> k;
  memset(sum, 0, sizeof(sum));
  for (int i = 0; i < n; i++) cin >> p[0][i];
  for (int i = 0; i < n; i++) {
    cin >> sum[0][i];
    low[0][i] = sum[0][i];
  }
  for (int i = 1; i < 37; i++) {
    for (int j = 0; j < n; j++) {
      p[i][j] = p[i - 1][p[i - 1][j]];
      low[i][j] = min(low[i - 1][j], low[i - 1][p[i - 1][j]]);
      sum[i][j] = (sum[i - 1][j] + sum[i - 1][p[i - 1][j]]);
    }
  }
  for (int i = 0; i < n; i++) {
    long long int mi = inf;
    long long int ss = 0;
    long long int val = k;
    long long int pos = i;
    for (int j = 35; j >= 0; j--) {
      if ((1ll << j) <= val) {
        ss = ss + sum[j][pos];
        mi = std::min(mi, low[j][pos]);
        pos = p[j][pos];
        val = val - (1ll << j);
      }
    }
    cout << ss << ' ' << mi << '\n';
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) solve();
}
