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;
int anc[100001][41], minn[100001][41];
long long sum[100001][41];
int n;
long long k;
vector<int> need;
long long mypow(int x) {
  long long ret = 1;
  for (int i = 1; i <= x; i++) ret *= 2;
  return ret;
}
int main() {
  int i, j;
  scanf("%d%I64d", &n, &k);
  for (i = 0; i <= n - 1; i++) scanf("%d", &anc[i][0]);
  for (i = 0; i <= n - 1; i++) {
    scanf("%d", &minn[i][0]);
    sum[i][0] = minn[i][0];
  }
  for (j = 1; j <= 40; j++)
    for (i = 0; i <= n - 1; i++) {
      anc[i][j] = anc[anc[i][j - 1]][j - 1];
      sum[i][j] = sum[i][j - 1] + sum[anc[i][j - 1]][j - 1];
      minn[i][j] = min(minn[i][j - 1], minn[anc[i][j - 1]][j - 1]);
    }
  long long t = k;
  int u = 0;
  while (mypow(u) <= t) u++;
  u--;
  while (t != 0) {
    need.push_back(u);
    t -= mypow(u);
    while (u >= 0 && mypow(u) > t) u--;
  }
  for (i = 0; i <= n - 1; i++) {
    long long ans1 = 0, ans2 = 1000000000000000000;
    int p = i;
    for (j = 0; j < need.size(); j++) {
      ans1 += sum[p][need[j]];
      ans2 = min(ans2, ((long long)minn[p][need[j]]));
      p = anc[p][need[j]];
    }
    printf("%I64d %I64d\n", ans1, ans2);
  }
  return 0;
}