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;
struct __io_dev {
  __io_dev(const bool& __fastio = false) {
    if (__fastio) ios_base::sync_with_stdio(false), cin.tie(nullptr);
    srand(time(nullptr));
    if (!string("").empty())
      freopen(
          ""
          ".in",
          "r", stdin),
          freopen(
              ""
              ".out",
              "w", stdout);
  }
  ~__io_dev() { fprintf(stderr, "%.6f ms\n", 1e3 * clock() / CLOCKS_PER_SEC); }
} __io(false);
const long long inf = (long long)1e+9 + 7ll;
const long long linf = (long long)1e+18 + 7ll;
const long double eps = (long double)1e-9;
const long double pi = acosl((long double)-1.0);
const int alph = 26;
const int maxs = 512l;
static char buff[(int)2e6 + 17];
long long __p[3] = {29ll, 31ll, 33ll};
long long __mod[3] = {inf, inf + 2ll, 14881337ll};
const int maxn = (int)1e5 + 17;
int n;
long long k;
int p[maxn][34];
long long sum[maxn][34], mn[maxn][34];
int main() {
  scanf("%d %lld", &n, &k);
  for (long long i = 0ll; i < (long long)n; ++i) scanf("%d", &p[i][0]);
  for (long long i = 0ll; i < (long long)n; ++i)
    scanf("%lld", &sum[i][0]), mn[i][0] = sum[i][0];
  for (long long j = (long long)1; j < (long long)34; ++j)
    for (long long i = 0ll; i < (long long)n; ++i) {
      p[i][j] = p[p[i][j - 1]][j - 1];
      sum[i][j] = sum[i][j - 1] + sum[p[i][j - 1]][j - 1];
      mn[i][j] = min(mn[i][j - 1], mn[p[i][j - 1]][j - 1]);
    }
  for (long long i = 0ll; i < (long long)n; ++i) {
    int v = i;
    long long curk = k, ansmin = linf, anssum = 0ll;
    for (int j = 33; j >= 0; --j)
      if ((1ll << j) <= curk) {
        ansmin = min(ansmin, mn[v][j]);
        anssum += sum[v][j];
        v = p[v][j];
        curk -= (1ll << j);
      }
    printf("%lld %lld\n", anssum, ansmin);
  }
  return 0;
}
