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 maxN = 1e5 + 7, MOD = 1e9 + 7, INF = 1e10, N = 5000 + 7;
long long n, k, cur, dp[maxN][50], d[maxN][50], sum[maxN][50], f[maxN], g[maxN],
    c[50], mini, ss;
vector<long long> q;
bool v[50], help, vis[maxN];
void dfs(int s) {
  vis[s] = true;
  if (!vis[f[s]]) dfs(f[s]);
  q.push_back(s);
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(false);
  c[0] = 1;
  for (long long i = 1; i <= 34; i++) c[i] = (long long)c[i - 1] * (long long)2;
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> f[i], f[i]++;
  for (long long i = 1; i <= n; i++) cin >> g[i];
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) help = false, dfs(i);
  }
  reverse(q.begin(), q.end());
  for (long long i = 1; i <= n; i++)
    dp[i][0] = f[i], d[i][0] = g[i], sum[i][0] = g[i];
  for (long long j = 1; j <= 34; j++) {
    for (int i = n - 1; i >= 0; i--)
      dp[q[i]][j] = dp[dp[q[i]][j - 1]][j - 1],
      sum[q[i]][j] = sum[q[i]][j - 1] + sum[dp[q[i]][j - 1]][j - 1],
      d[q[i]][j] = min(d[q[i]][j - 1], d[dp[q[i]][j - 1]][j - 1]);
  }
  for (int i = 34; i >= 0; i--) {
    if (k >= c[i]) v[i] = true, k -= c[i];
  }
  for (long long i = 1; i <= n; i++) {
    mini = INF;
    ss = 0;
    cur = i;
    for (int j = 34; j >= 0; j--) {
      if (v[j]) {
        ss += sum[cur][j];
        mini = min(mini, d[cur][j]);
        cur = dp[cur][j];
      }
    }
    cout << ss << " " << mini << "\n";
  }
  return cout << "\n", 0;
}
