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 maxn = 1e5 + 5;
const int mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
struct graph {
  int head[maxn], nxt[maxn << 1], to[maxn << 1], w[maxn << 1], sz;
  void init() { memset(head, -1, sizeof(head)); }
  graph() { init(); }
  void push(int a, int b, int c) {
    nxt[sz] = head[a], to[sz] = b, w[sz] = c, head[a] = sz++;
  }
  int& operator[](const int a) { return to[a]; }
};
long long dist[maxn][40], fi[maxn][40], mi[maxn][40];
int to[maxn], w[maxn];
int main() {
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) scanf("%d", &to[i]);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &w[i]);
    dist[i][0] = w[i];
    mi[i][0] = w[i];
    fi[i][0] = to[i];
  }
  for (int i = 1; i <= 39; ++i) {
    for (int j = 0; j < n; ++j) {
      dist[j][i] = dist[j][i - 1] + dist[fi[j][i - 1]][i - 1];
      fi[j][i] = fi[fi[j][i - 1]][i - 1];
      mi[j][i] = min(mi[j][i - 1], mi[fi[j][i - 1]][i - 1]);
    }
  }
  for (int i = 0; i < n; ++i) {
    long long mix = 1e18;
    long long ans = 0;
    long long res = k;
    int now = i;
    for (int j = 39; j >= 0; --j) {
      if (res >= (1ll << j)) {
        ans += dist[now][j];
        mix = min(mix, mi[now][j]);
        now = fi[now][j];
        res -= (1ll << j);
      }
    }
    printf("%lld %lld\n", ans, mix);
  }
  return 0;
}