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;
inline int rd() {
  register int ans = 0, flag = 1;
  register char c = getchar();
  for (; c > '9' || c < '0'; c = getchar())
    if (c == '-') flag = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) ans = ans * 10 - '0' + c;
  return ans * flag;
}
inline long long rdll() {
  register long long ans = 0, flag = 1;
  register char c = getchar();
  for (; c > '9' || c < '0'; c = getchar())
    if (c == '-') flag = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) ans = ans * 10 - '0' + c;
  return ans * flag;
}
const int MAXN = 1e5 + 10;
const int MAXLOGN = 50;
long long LOGN;
int to[MAXN], val[MAXN];
long long pt[MAXLOGN];
int n;
long long k;
int f[MAXN][MAXLOGN + 1];
long long m[MAXN][MAXLOGN + 1];
long long s[MAXN][MAXLOGN + 1];
int main() {
  n = rd();
  k = rdll();
  LOGN = 35;
  for (register int i = 1; i <= n; ++i) {
    to[i] = rd() + 1;
    f[i][0] = to[i];
  }
  for (register int i = 1; i <= n; ++i) {
    val[i] = rd();
    m[i][0] = val[i];
    s[i][0] = val[i];
  }
  pt[0] = 1;
  for (register int j = 1; j <= LOGN; ++j) {
    pt[j] = pt[j - 1] * 2;
    for (register int i = 1; i <= n; ++i) {
      f[i][j] = f[f[i][j - 1]][j - 1];
      m[i][j] = min(m[f[i][j - 1]][j - 1], m[i][j - 1]);
      s[i][j] = s[f[i][j - 1]][j - 1] + s[i][j - 1];
    }
  }
  for (register int i = 1; i <= n; ++i) {
    long long t = k;
    long long x = i;
    long long ans = INT_MAX;
    long long sum = 0;
    for (register int j = LOGN; j >= 0; --j) {
      if (pt[j] <= t) {
        t -= pt[j];
        sum += s[x][j];
        ans = min(ans, m[x][j]);
        x = f[x][j];
      }
    }
    printf("%lld %lld\n", sum, ans);
  }
  return 0;
}