Problem: Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.

The dice has m faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the m-th face contains m dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability <image>. Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice n times.

Input

A single line contains two integers m and n (1 ≤ m, n ≤ 105).

Output

Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10  - 4.

Examples

Input

6 1


Output

3.500000000000


Input

6 3


Output

4.958333333333


Input

2 2


Output

1.750000000000

Note

Consider the third test example. If you've made two tosses:

  1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 
  2. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 
  3. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 
  4. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2. 



The probability of each outcome is 0.25, that is expectation equals to: 

<image>

You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void sor(vector<long long> &x) { sort(x.begin(), x.end()); }
void rev(vector<long long> &x) { reverse(x.begin(), x.end()); }
long long gcd(long long x, long long y) {
  if (y == 0) {
    return x;
  }
  return gcd(y, x % y);
}
long long max(long long x, long long y) {
  if (x > y) {
    return x;
  } else {
    return y;
  }
}
long long min(long long x, long long y) {
  if (x > y) {
    return y;
  } else {
    return x;
  }
}
void primeFactors(long long n, vector<long long> &vec) {
  while (n % 2 == 0) {
    vec.push_back(2);
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      vec.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) vec.push_back(n);
}
long long lcm(long long x, long long y) { return x * y / gcd(x, y); }
long long g(long long x) {
  for (long long i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) {
      return i;
    }
  }
  return 0;
}
long long bina(vector<long long> a, long long n, long long m) {
  long long start = 0, end = n - 1;
  long long c = 0;
  if (m >= a[n - 1]) {
    return n;
  }
  while (start <= end) {
    long long mid = (start + end) / 2;
    if (m >= a[mid]) {
      c = mid + 1;
      start = mid + 1;
    } else {
      end = mid - 1;
    }
  }
  return c;
}
long long binarySearch(vector<long long> arr, long long l, long long r,
                       long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return arr[mid];
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
int subArraylen(long long arr[], long long n, long long K) {
  unordered_map<int, int> mp;
  mp[arr[0]] = 0;
  for (int i = 1; i < n; i++) {
    arr[i] = arr[i] + arr[i - 1];
    mp[arr[i]] = i;
  }
  int len = INT_MAX;
  for (int i = 0; i < n; i++) {
    if (arr[i] < K)
      continue;
    else {
      int x = arr[i] - K;
      if (x == 0) len = min(len, i);
      if (mp.find(x) == mp.end())
        continue;
      else {
        len = min(len, i - mp[x]);
      }
    }
  }
  return len;
}
long long prime(long long n) {
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      return i;
    }
  }
  return 0;
}
long long nextPowerOf2(long long n) {
  long long p = 1;
  if (n && !(n & (n - 1))) return n;
  while (p < n) p <<= 1;
  return p;
}
long long countBits(long long number) { return (long long)log2(number) + 1; }
long long comb(long long c, long long n) {
  long long num = 1, d = 1;
  for (long long i = c; i > max(n, c - n); i--) {
    num *= i;
  }
  for (long long i = 2; i <= min(n, c - n); i++) {
    d *= i;
  }
  return (num / d);
}
long long countDivisors(long long n) {
  long long cnt = 0;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return cnt;
}
int getCount(string a, string b) {
  if (b.length() % a.length() != 0) return -1;
  int count = b.length() / a.length();
  string str = "";
  for (int i = 0; i < count; i++) {
    str = str + a;
  }
  if (str == b) return count;
  return -1;
}
long long fact(long long n) {
  long long res = 1;
  for (int i = 2; i <= n; i++) res = res * i;
  return res;
}
long long nCr(long long n, long long r) {
  return fact(n) / (fact(r) * fact(n - r));
}
long long no(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long x = no(a, b / 2);
  if (b % 2) {
    return x * x * a;
  } else {
    return x * x;
  }
}
int countFreq(string &pat, string &txt) {
  int M = pat.length();
  int N = txt.length();
  int res = 0;
  for (int i = 0; i <= N - M; i++) {
    int j;
    for (j = 0; j < M; j++)
      if (txt[i + j] != pat[j]) break;
    if (j == M) {
      res++;
      j = 0;
    }
  }
  return res;
}
int highestPowerof2(int n) {
  int res = 0;
  for (int i = n; i >= 1; i--) {
    if ((i & (i - 1)) == 0) {
      res = i;
      break;
    }
  }
  return res;
}
vector<long long> gr[1000001];
long long aa[1000001];
long long bb[1000001];
void solve() {
  long long m, n;
  cin >> m >> n;
  double ans = 0;
  long long s = 0;
  long long nn = n;
  for (long double i = 0; i < m; i++) {
    double x = (i + 1) * (pow(double(i + 1) / m, n) - pow(double(i) / m, n));
    ans += x;
  }
  cout << fixed << setprecision(10) << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long t = 1;
  for (long long tt = 0; tt < t; ++tt) {
    solve();
  }
  return 0;
}