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 N = 100000;
const int LOGK = 41;
int par[LOGK][N];
long long sum[LOGK][N];
int mini[LOGK][N];
int main() {
  int n;
  long long k;
  scanf("%d %lld", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &par[0][i]);
  for (int i = 0; i < n; i++) scanf("%d", &sum[0][i]), mini[0][i] = sum[0][i];
  for (int i = 1; i < LOGK; i++) {
    for (int j = 0; j < n; j++) {
      par[i][j] = par[i - 1][par[i - 1][j]];
      sum[i][j] = sum[i - 1][j] + sum[i - 1][par[i - 1][j]];
      mini[i][j] = min(mini[i - 1][j], mini[i - 1][par[i - 1][j]]);
    }
  }
  for (int i = 0; i < n; i++) {
    long long s = 0;
    int m = 1000000000;
    long long _k = k;
    int cur = i;
    for (int j = 0; _k; _k >>= 1, j++) {
      if (_k & 1) {
        s += sum[j][cur];
        m = min(m, mini[j][cur]);
        cur = par[j][cur];
      }
    }
    printf("%lld %d\n", s, m);
  }
  return 0;
}
