Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int inf = 1e9;
const long long mod = 1000000007;
long long f[maxn], w[maxn];
long long a1[maxn], a2[maxn];
int cur[maxn];
long long sm[maxn][40], mi[maxn][40], nxt[maxn][40];
int main() {
  long long n, k;
  scanf("%I64d%I64d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%I64d", &f[i]);
  for (int i = 0; i < n; i++) scanf("%I64d", &w[i]);
  for (int i = 0; i < n; i++) {
    sm[i][0] = w[i];
    mi[i][0] = w[i];
    nxt[i][0] = f[i];
    cur[i] = i;
  }
  memset(a1, 0, sizeof(a1));
  memset(a2, 0x3f, sizeof(a2));
  for (long long _ = 1;; _++) {
    for (int i = 0; i < n; i++) {
      if (k & 1) {
        a1[i] += sm[cur[i]][_ - 1];
        a2[i] = min(a2[i], mi[cur[i]][_ - 1]);
        cur[i] = nxt[cur[i]][_ - 1];
      }
      int la = nxt[i][_ - 1];
      sm[i][_] = sm[i][_ - 1] + sm[la][_ - 1];
      mi[i][_] = min(mi[i][_ - 1], mi[la][_ - 1]);
      nxt[i][_] = nxt[la][_ - 1];
    }
    k >>= (long long)1;
    if (!k) break;
  }
  for (int i = 0; i < n; i++) cout << a1[i] << ' ' << a2[i] << endl;
  return 0;
}