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 maxp = 40;
const int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
int n, f[maxn], to[maxn][maxp];
long long k, s[maxn][maxp];
int w[maxn], minv[maxn][maxp];
long long si;
int mi;
void init() {
  for (int i = 0; i < n; i++) {
    to[i][0] = f[i];
    minv[i][0] = s[i][0] = w[i];
  }
  for (long long j = 1; (1ll << j) <= k; j++) {
    for (int i = 0; i < n; i++) {
      to[i][j] = to[to[i][j - 1]][j - 1];
      minv[i][j] = min(minv[i][j - 1], minv[to[i][j - 1]][j - 1]);
      s[i][j] = s[i][j - 1] + s[to[i][j - 1]][j - 1];
    }
  }
}
inline int lowbit(int x) { return x & -x; }
void query(int u, long long k) {
  mi = inf;
  si = 0;
  int cnt = 0;
  while (k) {
    while (!(k & 1)) {
      k >>= 1;
      cnt++;
    }
    mi = min(mi, minv[u][cnt]);
    si += s[u][cnt];
    u = to[u][cnt];
    k ^= 1;
  }
}
int main() {
  ios::sync_with_stdio("false");
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> f[i];
  for (int i = 0; i < n; i++) cin >> w[i];
  init();
  for (int i = 0; i < n; i++) {
    query(i, k);
    cout << si << " " << mi << endl;
  }
  return 0;
}
