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;
template <class T>
inline T lowbit(T x) {
  return x & (-x);
}
const long long INF = 1e18;
const int MAXN = 1e5 + 5;
const int BIT = 35;
int nxt[MAXN][BIT];
long long cost[MAXN][BIT], min_cost[MAXN][BIT];
int main(void) {
  int n;
  long long k;
  scanf("%d%lld", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &nxt[i][0]);
  }
  for (int i = 0; i < n; i++) {
    scanf("%lld", &cost[i][0]);
    min_cost[i][0] = cost[i][0];
  }
  for (int i = 1; i < BIT; i++) {
    for (int j = 0; j < n; j++) {
      nxt[j][i] = nxt[nxt[j][i - 1]][i - 1];
      cost[j][i] = cost[j][i - 1] + cost[nxt[j][i - 1]][i - 1];
      min_cost[j][i] = min(min_cost[j][i - 1], min_cost[nxt[j][i - 1]][i - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    long long s = 0, Min = INF;
    int u = i;
    for (int i = 0; i < BIT; i++) {
      if ((k >> i) & 1) {
        s += cost[u][i];
        Min = min(Min, min_cost[u][i]);
        u = nxt[u][i];
      }
    }
    printf("%lld %lld\n", s, Min);
  }
  return 0;
}