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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const long double PI = acosl((long double)-1.0);
const long long LINF = 1e18;
const int MOD = 1000000007;
const int LOG = 34;
const int MAXN = 200005;
int n;
long long k;
int f[LOG][MAXN];
int used[MAXN];
int c[MAXN];
int w[MAXN];
long long s[LOG][MAXN];
int mn[LOG][MAXN];
void init() {
  scanf("%d%I64d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &f[0][i]);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &w[i]);
  }
}
void solve() {
  init();
  for (int i = 0; i < n; i++) {
    s[0][i] = w[i];
    mn[0][i] = w[i];
  }
  for (int h = 1; h < LOG; h++) {
    for (int i = 0; i <= n; i++) {
      f[h][i] = f[h - 1][f[h - 1][i]];
      s[h][i] = s[h - 1][i] + s[h - 1][f[h - 1][i]];
      mn[h][i] = min(mn[h - 1][i], mn[h - 1][f[h - 1][i]]);
    }
  }
  for (int i = 0; i < n; i++) {
    int v = i;
    long long sum = 0;
    int m = ((int)(1e9) + 1337);
    for (int h = 0; (1LL << h) <= k; h++) {
      if (k & (1LL << h)) {
        sum += s[h][v];
        m = min(mn[h][v], m);
        v = f[h][v];
      }
    }
    printf("%I64d %d\n", sum, m);
  }
}
void precalc() {}
int main() {
  srand(707);
  precalc();
  int tests = 1;
  for (int i = 1; i <= tests; i++) {
    solve();
  }
  return 0;
}
