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:102400000,102400000")
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-4;
const long long mod = 1e9 + 7;
const unsigned long long mx = 133333331;
inline void RI(int &x) {
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0';
}
int pre[100005][40];
int w[100005][40];
long long s[100005][40];
long long ans1;
int ans2;
void solve(int u, long long k) {
  if (k == 0) {
    return;
  }
  int t = 0;
  while ((1LL << t) <= k) t++;
  t--;
  ans1 += s[u][t];
  ans2 = min(ans2, w[u][t]);
  solve(pre[u][t], k - (1LL << t));
}
int main() {
  int n;
  long long k;
  while (cin >> n >> k) {
    for (int i = 0; i < n; i++) scanf("%d", &pre[i][0]);
    for (int i = 0; i < n; i++) scanf("%d", &w[i][0]), s[i][0] = w[i][0];
    for (int i = 1; (1LL << i) <= k; i++) {
      for (int j = 0; j < n; j++) {
        pre[j][i] = pre[pre[j][i - 1]][i - 1];
        w[j][i] = min(w[j][i - 1], w[pre[j][i - 1]][i - 1]);
        s[j][i] = s[j][i - 1] + s[pre[j][i - 1]][i - 1];
      }
    }
    for (int i = 0; i < n; i++) {
      ans1 = 0;
      ans2 = INF;
      solve(i, k);
      printf("%I64d %d\n", ans1, ans2);
    }
  }
  return 0;
}
