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 mx = 1e5 + 10;
const double eps = 1e-9;
const long long inf = 1e18;
long long to[mx][40], par[mx];
long long mn[mx][40], sum[mx][40], w[mx], a[mx];
int main() {
  long long n, k;
  scanf("%lld%lld", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%lld", a + i);
  }
  for (int i = 0; i < n; i++) {
    scanf("%lld", w + i);
  }
  for (int i = 0; i < n; i++) {
    sum[i][0] = w[i];
    mn[i][0] = w[i];
    to[i][0] = a[i];
  }
  for (int lg = 1; lg < 35; lg++) {
    for (int i = 0; i < n; i++) {
      to[i][lg] = to[to[i][lg - 1]][lg - 1];
      sum[i][lg] = sum[i][lg - 1] + sum[to[i][lg - 1]][lg - 1];
      mn[i][lg] = min(mn[i][lg - 1], mn[to[i][lg - 1]][lg - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    int ii = i;
    long long sumans = 0, mnn = inf;
    for (int lg = 0; lg < 35; lg++) {
      if (k & (1ll << lg)) {
        mnn = min(mnn, 1ll * mn[ii][lg]);
        sumans += sum[ii][lg];
        ii = to[ii][lg];
      }
    }
    printf("%lld %lld\n", sumans, mnn);
  }
}
