#include <bits/stdc++.h>
using namespace std;
int break_point() {
  char c;
  while ((c = getchar()) != '\n')
    ;
  return 0;
}
template <typename T>
void read_integer(T &r) {
  bool sign = 0;
  r = 0;
  char c;
  while (1) {
    c = getchar();
    if (c == '-') {
      sign = 1;
      break;
    }
    if (c != ' ' && c != '\n') {
      r = c - '0';
      break;
    }
  }
  while (1) {
    c = getchar();
    if (c == ' ' || c == '\n') break;
    r = r * 10 + (c - '0');
  }
  if (sign) r = -r;
}
long long binpowmod(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  long long c = binpowmod(a, b >> 1, mod);
  return (((c * c) % mod) * (b & 1 ? a : 1)) % mod;
}
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long c = binpow(a, b >> 1);
  return c * c * (b & 1 ? a : 1);
}
inline int getbit(int x, int b) { return (x >> b) & 1; }
inline long long getbit(long long x, int b) { return (x >> b) & 1; }
inline unsigned long long getbit(unsigned long long x, int b) {
  return (x >> b) & 1;
}
inline int setbit(int x, int b) { return x | (1 << b); }
inline void _setbit(int &x, int b) { x = setbit(x, b); }
inline long long setbit(long long x, int b) { return x | (1ll << b); }
inline void _setbit(long long &x, int b) { x = setbit(x, b); }
inline int unsetbit(int x, int b) { return x & (INT_MAX - (1 << b)); }
inline void _unsetbit(int &x, int b) { x = unsetbit(x, b); }
inline int countbit(int x) {
  x = x - ((x >> 1) & 0x55555555);
  x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
  return ((x + (x >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
}
inline long long countbit(long long x) {
  int p1 = (x >> 32) & ((1ll << 32) - 1);
  int p2 = x & ((1ll << 32) - 1);
  return countbit(p1) + countbit(p2);
}
template <typename T>
void printbit(T x, int len) {
  for (int i = len - 1; i >= 0; i--) printf("%d", getbit(x, i));
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
template <typename A, typename B>
ostream &operator<<(ostream &stream, const pair<A, B> &p) {
  stream << "{" << p.first << "," << p.second << "}";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const vector<A> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A, typename B>
ostream &operator<<(ostream &stream, const map<A, B> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const set<A> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const stack<A> &v) {
  stack<A> st = v;
  stream << "[";
  while (!st.empty()) {
    stream << st.top() << " ";
    st.pop();
  }
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const priority_queue<A> &v) {
  priority_queue<A> q = v;
  stream << "[";
  while (!q.empty()) {
    stream << q.top() << " ";
    q.pop();
  }
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const queue<A> &v) {
  queue<A> q = v;
  stream << "[";
  while (!q.empty()) {
    stream << q.front() << " ";
    q.pop();
  }
  stream << "]";
  return stream;
}
void run();
int main() {
  srand(time(NULL));
  do {
    run();
    if (0) {
      0 ? printf("-------------------------------\n") : 0;
      0 ? printf("-------------------------------\n") : 0;
      0 ? break_point() : 0;
    }
  } while (0);
  return 0;
}
const int mod[3] = {int(1e9 + 7), int(1e9 + 9), int(1e9 + 21)};
int a[300005];
void run() {
  int n, k;
  scanf("%d%d", &n, &k);
  long long curpow2[3] = {1, 1, 1};
  long long val[3] = {0, 0, 0};
  for (int i = 0; i <= n; ++i) {
    scanf("%d", &a[i]);
    for (int j = 0; j < 3; j++) {
      val[j] = (val[j] + curpow2[j] * a[i]) % mod[j];
      curpow2[j] = (curpow2[j] * 2) % mod[j];
    }
  }
  int res = 0;
  for (int i = 0; i <= n; i++) {
    long long cur[3], currev[3];
    for (int j = 0; j < 3; j++) {
      cur[j] = val[j] - (a[i] * binpowmod(2, i, mod[j])) % mod[j];
      while (cur[j] < 0) cur[j] += mod[j];
      cur[j] =
          (cur[j] * binpowmod(binpowmod(2, i, mod[j]), mod[j] - 2, mod[j])) %
          mod[j];
      currev[j] = cur[j] - mod[j];
    }
    bool f[2] = {true, true};
    for (int j = 1; j < 3; j++) {
      f[0] &= (cur[j] == cur[j - 1]);
      f[1] &= (currev[j] == currev[j - 1]);
    }
    if (f[0] && cur[0] == 0 && i == n) continue;
    res += (f[0] && cur[0] <= k);
    res += (f[1] && currev[0] >= -k);
  }
  printf("%d", res);
  putchar('\n');
}

### Prompt

Create a solution in CPP for the following problem:

Limak is a little polar bear. He doesn't have many toys and thus he often plays with polynomials.

He considers a polynomial valid if its degree is n and its coefficients are integers not exceeding k by the absolute value. More formally:

Let a0, a1, ..., an denote the coefficients, so <image>. Then, a polynomial P(x) is valid if all the following conditions are satisfied:

  * ai is integer for every i; 
  * |ai| ≤ k for every i; 
  * an ≠ 0. 



Limak has recently got a valid polynomial P with coefficients a0, a1, a2, ..., an. He noticed that P(2) ≠ 0 and he wants to change it. He is going to change one coefficient to get a valid polynomial Q of degree n that Q(2) = 0. Count the number of ways to do so. You should count two ways as a distinct if coefficients of target polynoms differ.

Input

The first line contains two integers n and k (1 ≤ n ≤ 200 000, 1 ≤ k ≤ 109) — the degree of the polynomial and the limit for absolute values of coefficients.

The second line contains n + 1 integers a0, a1, ..., an (|ai| ≤ k, an ≠ 0) — describing a valid polynomial <image>. It's guaranteed that P(2) ≠ 0.

Output

Print the number of ways to change one coefficient to get a valid polynomial Q that Q(2) = 0.

Examples

Input

3 1000000000
10 -9 -3 5


Output

3


Input

3 12
10 -9 -3 5


Output

2


Input

2 20
14 -7 19


Output

0

Note

In the first sample, we are given a polynomial P(x) = 10 - 9x - 3x2 + 5x3.

Limak can change one coefficient in three ways:

  1. He can set a0 = - 10. Then he would get Q(x) = - 10 - 9x - 3x2 + 5x3 and indeed Q(2) = - 10 - 18 - 12 + 40 = 0. 
  2. Or he can set a2 = - 8. Then Q(x) = 10 - 9x - 8x2 + 5x3 and indeed Q(2) = 10 - 18 - 32 + 40 = 0. 
  3. Or he can set a1 = - 19. Then Q(x) = 10 - 19x - 3x2 + 5x3 and indeed Q(2) = 10 - 38 - 12 + 40 = 0. 



In the second sample, we are given the same polynomial. This time though, k is equal to 12 instead of 109. Two first of ways listed above are still valid but in the third way we would get |a1| > k what is not allowed. Thus, the answer is 2 this time.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int break_point() {
  char c;
  while ((c = getchar()) != '\n')
    ;
  return 0;
}
template <typename T>
void read_integer(T &r) {
  bool sign = 0;
  r = 0;
  char c;
  while (1) {
    c = getchar();
    if (c == '-') {
      sign = 1;
      break;
    }
    if (c != ' ' && c != '\n') {
      r = c - '0';
      break;
    }
  }
  while (1) {
    c = getchar();
    if (c == ' ' || c == '\n') break;
    r = r * 10 + (c - '0');
  }
  if (sign) r = -r;
}
long long binpowmod(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  long long c = binpowmod(a, b >> 1, mod);
  return (((c * c) % mod) * (b & 1 ? a : 1)) % mod;
}
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long c = binpow(a, b >> 1);
  return c * c * (b & 1 ? a : 1);
}
inline int getbit(int x, int b) { return (x >> b) & 1; }
inline long long getbit(long long x, int b) { return (x >> b) & 1; }
inline unsigned long long getbit(unsigned long long x, int b) {
  return (x >> b) & 1;
}
inline int setbit(int x, int b) { return x | (1 << b); }
inline void _setbit(int &x, int b) { x = setbit(x, b); }
inline long long setbit(long long x, int b) { return x | (1ll << b); }
inline void _setbit(long long &x, int b) { x = setbit(x, b); }
inline int unsetbit(int x, int b) { return x & (INT_MAX - (1 << b)); }
inline void _unsetbit(int &x, int b) { x = unsetbit(x, b); }
inline int countbit(int x) {
  x = x - ((x >> 1) & 0x55555555);
  x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
  return ((x + (x >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
}
inline long long countbit(long long x) {
  int p1 = (x >> 32) & ((1ll << 32) - 1);
  int p2 = x & ((1ll << 32) - 1);
  return countbit(p1) + countbit(p2);
}
template <typename T>
void printbit(T x, int len) {
  for (int i = len - 1; i >= 0; i--) printf("%d", getbit(x, i));
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
template <typename A, typename B>
ostream &operator<<(ostream &stream, const pair<A, B> &p) {
  stream << "{" << p.first << "," << p.second << "}";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const vector<A> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A, typename B>
ostream &operator<<(ostream &stream, const map<A, B> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const set<A> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const stack<A> &v) {
  stack<A> st = v;
  stream << "[";
  while (!st.empty()) {
    stream << st.top() << " ";
    st.pop();
  }
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const priority_queue<A> &v) {
  priority_queue<A> q = v;
  stream << "[";
  while (!q.empty()) {
    stream << q.top() << " ";
    q.pop();
  }
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const queue<A> &v) {
  queue<A> q = v;
  stream << "[";
  while (!q.empty()) {
    stream << q.front() << " ";
    q.pop();
  }
  stream << "]";
  return stream;
}
void run();
int main() {
  srand(time(NULL));
  do {
    run();
    if (0) {
      0 ? printf("-------------------------------\n") : 0;
      0 ? printf("-------------------------------\n") : 0;
      0 ? break_point() : 0;
    }
  } while (0);
  return 0;
}
const int mod[3] = {int(1e9 + 7), int(1e9 + 9), int(1e9 + 21)};
int a[300005];
void run() {
  int n, k;
  scanf("%d%d", &n, &k);
  long long curpow2[3] = {1, 1, 1};
  long long val[3] = {0, 0, 0};
  for (int i = 0; i <= n; ++i) {
    scanf("%d", &a[i]);
    for (int j = 0; j < 3; j++) {
      val[j] = (val[j] + curpow2[j] * a[i]) % mod[j];
      curpow2[j] = (curpow2[j] * 2) % mod[j];
    }
  }
  int res = 0;
  for (int i = 0; i <= n; i++) {
    long long cur[3], currev[3];
    for (int j = 0; j < 3; j++) {
      cur[j] = val[j] - (a[i] * binpowmod(2, i, mod[j])) % mod[j];
      while (cur[j] < 0) cur[j] += mod[j];
      cur[j] =
          (cur[j] * binpowmod(binpowmod(2, i, mod[j]), mod[j] - 2, mod[j])) %
          mod[j];
      currev[j] = cur[j] - mod[j];
    }
    bool f[2] = {true, true};
    for (int j = 1; j < 3; j++) {
      f[0] &= (cur[j] == cur[j - 1]);
      f[1] &= (currev[j] == currev[j - 1]);
    }
    if (f[0] && cur[0] == 0 && i == n) continue;
    res += (f[0] && cur[0] <= k);
    res += (f[1] && currev[0] >= -k);
  }
  printf("%d", res);
  putchar('\n');
}

```