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 long long NMAX = 100000, LOGK = 35;
long long S[LOGK][NMAX], M[LOGK][NMAX], D[LOGK][NMAX];
long long f[NMAX], w[NMAX];
int main() {
  long long n, k;
  scanf("%lld%lld", &n, &k);
  for (int i = 0; i < n; i++) scanf("%lld", f + i);
  for (int i = 0; i < n; i++) scanf("%lld", w + i);
  for (int i = 0; i < n; i++) S[0][i] = M[0][i] = w[i];
  for (int i = 0; i < n; i++) D[0][i] = f[i];
  for (int i = 1; i < LOGK; i++)
    for (int j = 0; j < n; j++) {
      S[i][j] = S[i - 1][j] + S[i - 1][D[i - 1][j]];
      M[i][j] = min(M[i - 1][j], M[i - 1][D[i - 1][j]]);
      D[i][j] = D[i - 1][D[i - 1][j]];
    }
  for (int i = 0; i < n; i++) {
    int it = i;
    long long s = 0, m = 0x7fFFffFFffFFffFFLL;
    for (int j = 0; j < LOGK; j++)
      if (k & (1LL << j)) {
        s += S[j][it];
        m = min(m, M[j][it]);
        it = D[j][it];
      }
    printf("%lld %lld\n", s, m);
  }
}
