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>
using namespace std;
const int maxn = 100005;
long long dp[maxn][35];
long long sum[maxn][35];
int f[maxn][35];
int n;
long long k;
void init() {
  for (int i = 0; i < maxn - 1; i++) {
    for (int j = 0; j < 35; j++) {
      dp[i][j] = 1e18;
    }
  }
}
void DP() {
  for (long long j = 1; (1LL << j) <= k; j++) {
    for (int i = 0; i < n; i++) {
      int p = f[i][j - 1];
      f[i][j] = f[p][j - 1];
      dp[i][j] = min(dp[i][j], min(dp[i][j - 1], dp[p][j - 1]));
      sum[i][j] += sum[i][j - 1] + sum[p][j - 1];
    }
  }
}
int main() {
  init();
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    f[i][0] = x;
  }
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    dp[i][0] = x;
    sum[i][0] = x;
  }
  DP();
  for (int i = 0; i < n; i++) {
    long long T = k;
    long long ans1 = 1e18;
    long long ans2 = 0;
    int pre = i;
    for (int j = 34; j >= 0; j--) {
      if ((1LL << j) <= T) {
        ans1 = min(ans1, dp[pre][j]);
        ans2 += sum[pre][j];
        pre = f[pre][j];
        T -= (1LL << j);
      }
      if (T == 0) break;
    }
    cout << ans2 << " " << ans1 << endl;
  }
}
