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 N = 1e5 + 5, L = 40, mod = 0;
long long f[N][L], mn[N][L], sum[N][L];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; ++i) cin >> f[i][0];
  for (long long i = 0; i < n; ++i) cin >> sum[i][0], mn[i][0] = sum[i][0];
  for (long long g = 1; g < L; ++g)
    for (long long v = 0; v < n; ++v)
      f[v][g] = f[f[v][g - 1]][g - 1],
      mn[v][g] = min(mn[v][g - 1], mn[f[v][g - 1]][g - 1]),
      sum[v][g] = sum[v][g - 1] + sum[f[v][g - 1]][g - 1];
  for (long long v = 0; v < n; ++v) {
    long long mres = 1e18, sres = 0, cur = v;
    for (long long i = L - 1; i >= 0; --i)
      if (k >> i & 1) {
        mres = min(mres, mn[cur][i]);
        sres += sum[cur][i];
        cur = f[cur][i];
      }
    cout << sres << ' ' << mres << '\n';
  }
}
