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:102400000000,102400000000")
const long long MOD = 998244353;
const int N = 1e5 + 15;
const int maxn = 1e5 + 15;
const int letter = 130;
const int INF = 2e9;
const double pi = acos(-1.0);
const double eps = 1e-6;
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, f[N][37], mn[N][37], maxm = 37;
long long k, sum[N][37];
int main() {
  scanf("%d%I64d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &f[i][0]);
  for (int i = 0; i < n; i++) scanf("%d", &mn[i][0]), sum[i][0] = mn[i][0];
  for (int j = 1; j < maxm; j++)
    for (int i = 0; i < n; i++) {
      f[i][j] = f[f[i][j - 1]][j - 1];
      sum[i][j] = sum[i][j - 1] + sum[f[i][j - 1]][j - 1];
      mn[i][j] = min(mn[i][j - 1], mn[f[i][j - 1]][j - 1]);
    }
  for (int i = 0; i < n; i++) {
    int min1 = INF, x = i;
    long long ans = 0;
    for (int j = maxm - 1; j >= 0; j--) {
      if ((1ll << j) & k) {
        min1 = min(min1, mn[x][j]);
        ans += sum[x][j];
        x = f[x][j];
      }
    }
    printf("%I64d %d\n", ans, min1);
  }
  return 0;
}
