Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const int SQRTN = 320;
const int LOGN = 40;
const double PI = acos(-1);
long long dp[N][LOGN], dp1[N][LOGN], dp2[N][LOGN], n, f[N], w[N], k;
void calc() {
  for (int i = 0; i < n; i++) dp[i][0] = f[i];
  for (int i = 1; i < LOGN; i++) {
    for (int j = 0; j < n; j++) {
      dp[j][i] = dp[dp[j][i - 1]][i - 1];
    }
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < LOGN; j++) dp2[i][j] = 2000000000;
  }
  for (int i = 0; i < n; i++) dp1[i][0] = dp2[i][0] = w[i];
  for (int i = 1; i < LOGN; i++) {
    for (int j = 0; j < n; j++) {
      dp1[j][i] = dp1[dp[j][i - 1]][i - 1] + dp1[j][i - 1];
    }
  }
  for (int i = 1; i < LOGN; i++) {
    for (int j = 0; j < n; j++) {
      dp2[j][i] = min(dp2[dp[j][i - 1]][i - 1], dp2[j][i - 1]);
    }
  }
}
long long get(long long n, long long k) {
  long long ret = 0;
  for (long long i = 36; i > -1; i--) {
    if (((1ll << i) & k)) {
      ret += dp1[n][i];
      n = dp[n][i];
      k ^= (1ll << i);
    }
  }
  return ret;
}
long long get2(long long n, long long k) {
  long long ret = 2000000000;
  for (int i = 36; i > -1; i--) {
    if ((1ll << i) & k) {
      ret = min(ret, dp2[n][i]);
      n = dp[n][i];
      k -= (1ll << i);
    }
  }
  return ret;
}
int main() {
  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);
  calc();
  for (int j = 0; j < n; j++) {
    printf("%lld %lld\n", get(j, k), get2(j, k));
  }
  return 0;
}