Problem: Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor. 

Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.

Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds. 

Your task is to help him solve this complicated task.

Input

The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.

Output

Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.

Examples

Input

1 0.50 1


Output

0.5


Input

1 0.50 4


Output

0.9375


Input

4 0.20 2


Output

0.4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
inline int lcm(int a, int b) { return (a * (b / gcd(a, b))); }
inline long long int bigmod(long long int a, long long int b, long long int m) {
  if (b == 0) return 1 % m;
  long long int x = bigmod(a, b / 2, m);
  x = (x * x) % m;
  if (b % 2 == 1) {
    x = (x * a) % m;
  }
  return x;
}
inline long long int phi(long long int n) {
  long long int ret = n;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
    }
    ret -= ret / i;
  }
  if (n > 1) {
    ret -= ret / n;
  }
  return ret;
}
long long int missing_number(long long int A[], long long int n) {
  long long int x1 = A[0], x2 = 1;
  for (int i = 1; i < n; i++) x1 ^= A[i];
  for (int i = 2; i <= n + 1; i++) x2 ^= i;
  return (x1 ^ x2);
}
bool isFibo(long long int n) {
  long long int A = (5 * n * n + 4);
  long long int B = (5 * n * n - 4);
  long long int C = sqrt(A);
  long long int D = sqrt(B);
  return (A == (C * C) || B == (D * D));
}
long long int classini(long long int n) { return (n & 1) ? -1 : 1; }
long long int factorial(int n) {
  return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}
long long int catanal(long long int n) {
  return (factorial(2 * n) / ((factorial(n + 1) * factorial(n))));
}
string orientation_3_ordered_point(vector<pair<int, int> > V) {
  long long int value =
      (V[1].second - V[0].second) * (V[2].first - V[1].first) -
      (V[1].first * V[0].first) * (V[2].second - V[1].second);
  string res;
  if (value == 0) res = "colinear";
  value == 1 ? res = "clockwise" : res = "counterclockwise";
  return res;
}
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int fx[] = {+0, +0, +1, -1, -1, +1, -1, +1};
int fy[] = {-1, +1, +0, +0, +1, +1, -1, -1};
string make_string(long long int num) {
  stringstream ss;
  ss << num;
  string s;
  ss >> s;
  return s;
}
long long int make_num(string s) {
  stringstream ss;
  long long int tmp;
  ss << s;
  ss >> tmp;
  return tmp;
}
int SET(int mask, int pos) { return (mask | (1 << pos)); }
int check(int mask, int pos) { return (mask & (1 << pos)); }
double dp[2100][2100];
int vis[2100][2100];
double n, p;
double t;
double ans_me(int cur, int tm) {
  if (cur == 0) return 0;
  if (tm == 0) return 0;
  if (vis[cur][tm] != 0) return dp[cur][tm];
  vis[cur][tm] = 1;
  double ans = 0;
  ans = p * (ans_me(cur - 1, tm - 1) + 1);
  ans += (1 - p) * (ans_me(cur, tm - 1));
  return dp[cur][tm] = ans;
}
int main() {
  cin >> n >> p >> t;
  printf("%.8f\n", ans_me(n, t));
  return 0;
}