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 maxN = 1e5 + 5;
const int maxLOG = 35;
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long sq(long long x) { return (x * x) % MOD; }
long long modP(long long a, long long b) {
  return (!b ? 1 : (sq(modP(a, b / 2)) * (b % 2 ? a : 1)) % MOD);
}
long long nxt[maxN][maxLOG], sum[maxN][maxLOG], mn[maxN][maxLOG];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, inp;
  cin >> n;
  long long k;
  cin >> k;
  for (int i = 0; i < n; i++) {
    cin >> inp;
    nxt[i][0] = inp;
  }
  for (int i = 0; i < n; i++) {
    cin >> inp;
    sum[i][0] = inp;
    mn[i][0] = inp;
  }
  for (int i = 1; i < maxLOG; i++) {
    for (int v = 0; v < n; v++) {
      nxt[v][i] = nxt[nxt[v][i - 1]][i - 1];
    }
  }
  for (int i = 1; i < maxLOG; i++) {
    for (int v = 0; v < n; v++) {
      sum[v][i] = sum[v][i - 1] + sum[nxt[v][i - 1]][i - 1];
      mn[v][i] = min(mn[v][i - 1], mn[nxt[v][i - 1]][i - 1]);
    }
  }
  for (int v = 0; v < n; v++) {
    long long u = v;
    long long s = 0;
    long long m = INF;
    for (long long i = maxLOG - 1; i > -1; i--) {
      if ((k >> i) % 2 == 1) {
        s += sum[u][i];
        m = min(m, mn[u][i]);
        u = nxt[u][i];
      }
    }
    cout << s << " " << m << "\n";
  }
  return 0;
}
