Problem: It's been long after the events of the previous problems, and Karen has now moved on from student life and is looking to relocate to a new neighborhood.

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The neighborhood consists of n houses in a straight line, labelled 1 to n from left to right, all an equal distance apart.

Everyone in this neighborhood loves peace and quiet. Because of this, whenever a new person moves into the neighborhood, he or she always chooses the house whose minimum distance to any occupied house is maximized. If there are multiple houses with the maximum possible minimum distance, he or she chooses the leftmost one.

Note that the first person to arrive always moves into house 1.

Karen is the k-th person to enter this neighborhood. If everyone, including herself, follows this rule, which house will she move into?

Input

The first and only line of input contains two integers, n and k (1 ≤ k ≤ n ≤ 1018), describing the number of houses in the neighborhood, and that Karen was the k-th person to move in, respectively.

Output

Output a single integer on a line by itself, the label of the house Karen will move into.

Examples

Input

6 4


Output

2


Input

39 3


Output

20

Note

In the first test case, there are 6 houses in the neighborhood, and Karen is the fourth person to move in:

  1. The first person moves into house 1. 
  2. The second person moves into house 6. 
  3. The third person moves into house 3. 
  4. The fourth person moves into house 2. 



In the second test case, there are 39 houses in the neighborhood, and Karen is the third person to move in:

  1. The first person moves into house 1. 
  2. The second person moves into house 39. 
  3. The third person moves into house 20. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(0.0) * 2.0;
const long double eps = 1e-10;
const int step[8][2] = {{-1, 0}, {0, 1}, {1, 0},  {0, -1},
                        {-1, 1}, {1, 1}, {1, -1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <typename t, typename t1>
t min1(t a, t1 b) {
  return a < b ? a : b;
}
template <typename t, typename... arg>
t min1(t a, arg... arr) {
  return min1(a, min1(arr...));
}
template <typename t, typename t1>
t max1(t a, t1 b) {
  return a > b ? a : b;
}
template <typename t, typename... arg>
t max1(t a, arg... arr) {
  return max1(a, max1(arr...));
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / max(abs1(a), abs1(b)) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename t>
inline int jud(t a, t b) {
  if (a < b) return -1;
  if (a == b) return 0;
  return 1;
}
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  if (na == 0) return 0;
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
template <class T>
inline T lowb(T num) {
  return num & (-num);
}
inline int bitnum(unsigned int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(unsigned long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitnum(long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitmaxl(unsigned int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(unsigned long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
inline int bitmaxl(long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
long long pow(long long n, long long m, long long mod = 0) {
  if (m < 0) return 0;
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
template <class t1, class t2>
inline void add(t1 &a, t2 b, int mod = -1) {
  if (mod == -1) mod = 1000000007;
  a += b;
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
}
template <class t>
void output1(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) cerr << arr[i] << ' ';
  cerr << endl;
}
template <class t>
void output2(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) output1(arr[i]);
}
long long n, k;
long long get(long long l, long long val) {
  map<long long, long long> dp;
  dp[l] = 1;
  long long ans = 0;
  for (auto it = dp.rbegin(); it != dp.rend(); it++) {
    if (it->first < val * 2 + 1) {
      break;
    }
    if (it->first == val * 2 + 1 || it->first == val * 2 + 2) {
      ans += it->second;
      if (it->first == 2) {
        ans += it->second;
      }
    }
    if (it->first <= 2) {
      continue;
    }
    dp[it->first / 2] += it->second;
    dp[it->first - 1 - it->first / 2] += it->second;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  if (k == 1) {
    cout << 1 << endl;
    return 0;
  }
  if (k == 2) {
    cout << n << endl;
    return 0;
  }
  k -= 2, n -= 2;
  map<long long, long long> dp, rdp;
  dp[n] = 1;
  long long cnt = 0, rval = 0;
  for (auto it = dp.rbegin(); cnt < k; it++) {
    rval = it->first - 1;
    rval /= 2;
    cnt += it->second;
    rdp[rval] += it->second;
    dp[it->first / 2] += it->second;
    dp[it->first - it->first / 2 - 1] += it->second;
  }
  cnt -= rdp[rval];
  k -= cnt;
  long long ans = 1;
  for (long long rl = n; (rl - 1) / 2 > rval;) {
    long long lleft = rl - rl / 2 - 1;
    long long lright = rl / 2;
    long long rcnt = get(lleft, rval);
    if (rcnt >= k) {
      rl = lleft;
    } else {
      k -= rcnt;
      rl = lright;
      ans += lleft + 1;
    }
  }
  if (k == 2)
    ans += 2;
  else
    ans += rval + 1;
  cout << ans << endl;
  return 0;
}