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 LOG = 40;
const int MAXN = 101000;
int n;
int p[MAXN];
int w[MAXN];
int go[LOG][MAXN];
long long sm[LOG][MAXN];
long long mn[LOG][MAXN];
long long k;
int main() {
  scanf("%d%lld", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", p + i);
  for (int i = 0; i < n; ++i) scanf("%d", w + i);
  for (int i = 0; i < n; ++i) {
    go[0][i] = p[i];
    sm[0][i] = w[i];
    mn[0][i] = w[i];
  }
  for (int i = 1; i < LOG; ++i) {
    for (int j = 0; j < n; ++j) {
      go[i][j] = go[i - 1][go[i - 1][j]];
      sm[i][j] = sm[i - 1][j] + sm[i - 1][go[i - 1][j]];
      mn[i][j] = min(mn[i - 1][j], mn[i - 1][go[i - 1][j]]);
    }
  }
  for (int i = 0; i < n; ++i) {
    long long mni = 1e9;
    long long sum = 0;
    long long nk = k;
    int now = i;
    for (int j = LOG - 1; j >= 0; --j) {
      if ((1ll << j) <= nk) {
        nk -= (1ll << j);
        sum += sm[j][now];
        mni = min(mni, mn[j][now]);
        now = go[j][now];
      }
    }
    printf("%lld %lld\n", sum, mni);
  }
  return 0;
}