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 MAXN = 1e5 + 10, MAXL = 40;
const long long INF = 4e18;
int n;
long long m;
int maxl;
int pai[MAXN][MAXL];
long long lessy[MAXN][MAXL];
long long sum[MAXN][MAXL];
void build() {
  maxl = log2(m) + 1;
  for (int l = 1; l <= maxl; l++) {
    for (int i = 0; i < n; i++) {
      pai[i][l] = pai[pai[i][l - 1]][l - 1];
      sum[i][l] = sum[pai[i][l - 1]][l - 1] + sum[i][l - 1];
      lessy[i][l] = min(lessy[pai[i][l - 1]][l - 1], lessy[i][l - 1]);
    }
  }
}
int main() {
  scanf("%d %lld", &n, &m);
  for (int i = 0, a; i < n; i++) scanf("%d", &a), pai[i][0] = a;
  long long b;
  for (int i = 0, b; i < n; i++) scanf("%lld", &b), lessy[i][0] = sum[i][0] = b;
  build();
  for (int i = 0; i < n; i++) {
    long long jump = m;
    int cur = i;
    long long lessy0 = INF, sum0 = 0;
    for (int l = maxl; jump && l >= 0; l--) {
      if ((1LL << l) <= jump) {
        lessy0 = min(lessy0, lessy[cur][l]);
        sum0 += sum[cur][l];
        cur = pai[cur][l];
        jump -= (1LL << l);
      }
    }
    printf("%lld %lld\n", sum0, lessy0);
  }
}