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 inf = 0x7fffffff;
const long long INF = 0x7fffffffffffffff;
template <typename A, typename B>
ostream &operator<<(ostream &s, const pair<A, B> &p) {
  return s << "(" << p.first << "," << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &s, const vector<T> &c) {
  s << "[ ";
  for (auto it : c) s << it << " ";
  s << "]";
  return s;
}
template <typename T>
ostream &operator<<(ostream &o, const set<T> &st) {
  o << "{";
  for (auto it = st.begin(); it != st.end(); it++)
    o << (it == st.begin() ? "" : ", ") << *it;
  return o << "}";
}
template <typename T1, typename T2>
ostream &operator<<(ostream &o, const map<T1, T2> &make_pair) {
  o << "{";
  for (auto it = make_pair.begin(); it != make_pair.end(); it++) {
    o << (it == make_pair.begin() ? "" : ", ") << it->first << ":"
      << it->second;
  }
  o << "}";
  return o;
}
int n;
long long k;
int mn[100005][40], nxt[100005][40];
long long sum[100005][40];
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) scanf("%d", &nxt[i][0]);
  for (int i = 0; i < n; i++) scanf("%lld", &sum[i][0]), mn[i][0] = sum[i][0];
  for (int i = 1; i < 40; i++) {
    for (int j = 0; j < n; j++) {
      nxt[j][i] = nxt[nxt[j][i - 1]][i - 1];
      sum[j][i] = sum[j][i - 1] + sum[nxt[j][i - 1]][i - 1];
      mn[j][i] = min(mn[j][i - 1], mn[nxt[j][i - 1]][i - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    long long kk = k;
    int m = 1e9;
    long long s = 0;
    int cur = i;
    for (int j = 39; kk; j--)
      if ((1LL << j) <= kk) {
        kk -= (1LL << j);
        s += sum[cur][j];
        m = min(m, mn[cur][j]);
        cur = nxt[cur][j];
      }
    assert(kk == 0);
    printf("%lld %d\n", s, m);
  }
}
