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 sa_next[34][100010];
int sa_min[34][100010];
long long sa_sum[34][100010];
void Solve() {
  int i, j, n, m;
  long long k;
  scanf("%d %I64d", &n, &k);
  for (int _i = 0; _i < (n); _i++) scanf("%d", &((sa_next[0])[_i]));
  for (i = 0; i < n; i++) {
    scanf("%d", &(sa_min[0][i]));
    sa_sum[0][i] = sa_min[0][i];
  }
  for (i = 1; i < 34; i++)
    for (j = 0; j < n; j++) {
      sa_next[i][j] = sa_next[i - 1][sa_next[i - 1][j]];
      sa_min[i][j] = min(sa_min[i - 1][j], sa_min[i - 1][sa_next[i - 1][j]]);
      sa_sum[i][j] = sa_sum[i - 1][j] + sa_sum[i - 1][sa_next[i - 1][j]];
    }
  for (i = 0; i < n; i++) {
    long long v = 1ll << (34 - 1);
    long long kk = k;
    int mn = 1000000000;
    long long sum = 0;
    int u = i;
    for (j = 34 - 1; j >= 0; j--, v >>= 1) {
      if (v <= kk) {
        kk -= v;
        sum += sa_sum[j][u];
        mn = min(mn, sa_min[j][u]);
        u = sa_next[j][u];
      }
    }
    printf("%I64d %d\n", sum, mn);
  }
}
int main() {
  Solve();
  return 0;
}