Note that the memory limit in this problem is lower than in others.

You have a vertical strip with n cells, numbered consecutively from 1 to n from top to bottom.

You also have a token that is initially placed in cell n. You will move the token up until it arrives at cell 1.

Let the token be in cell x > 1 at some moment. One shift of the token can have either of the following kinds: 

  * Subtraction: you choose an integer y between 1 and x-1, inclusive, and move the token from cell x to cell x - y. 
  * Floored division: you choose an integer z between 2 and x, inclusive, and move the token from cell x to cell ⌊ x/z ⌋ (x divided by z rounded down). 



Find the number of ways to move the token from cell n to cell 1 using one or more shifts, and print it modulo m. Note that if there are several ways to move the token from one cell to another in one shift, all these ways are considered distinct (check example explanation for a better understanding).

Input

The only line contains two integers n and m (2 ≤ n ≤ 4 ⋅ 10^6; 10^8 < m < 10^9; m is a prime number) — the length of the strip and the modulo.

Output

Print the number of ways to move the token from cell n to cell 1, modulo m.

Examples

Input


3 998244353


Output


5


Input


5 998244353


Output


25


Input


42 998244353


Output


793019428


Input


787788 100000007


Output


94810539

Note

In the first test, there are three ways to move the token from cell 3 to cell 1 in one shift: using subtraction of y = 2, or using division by z = 2 or z = 3.

There are also two ways to move the token from cell 3 to cell 1 via cell 2: first subtract y = 1, and then either subtract y = 1 again or divide by z = 2.

Therefore, there are five ways in total.
#include <bits/stdc++.h>
namespace IO {
const int SIZE = (1 << 20) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = obuf + SIZE - 1;
char _st[55];
int _qr = 0;
inline char gc() {
  return (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, SIZE, stdin),
          (iS == iT ? EOF : *iS++) : *iS++);
}
inline void qread() {}
template <class T1, class... T2>
inline void qread(T1 &IEE, T2 &...ls) {
  register T1 __ = 0, ___ = 1;
  register char ch;
  while (!isdigit(ch = gc())) ___ = (ch == '-') ? -___ : ___;
  do {
    __ = (__ << 1) + (__ << 3) + (ch ^ 48);
  } while (isdigit(ch = gc()));
  __ *= ___;
  IEE = __;
  qread(ls...);
  return;
}
template <class T>
inline void QreadArr(T Begin, T End) {
  while (Begin != End) {
    qread(*Begin);
    ++Begin;
  }
}
inline void flush() {
  fwrite(obuf, 1, oS - obuf, stdout);
  oS = obuf;
  return;
}
inline void putc_(char _x) {
  *oS++ = _x;
  if (oS == oT) flush();
}
inline void qwrite() {}
template <class T1, class... T2>
inline void qwrite(T1 IEE, T2... ls) {
  if (!IEE) putc_('0');
  if (IEE < 0) putc_('-'), IEE = -IEE;
  while (IEE) _st[++_qr] = IEE % 10 + '0', IEE /= 10;
  while (_qr) putc_(_st[_qr--]);
  qwrite(ls...);
  return;
}
template <class T>
inline void WriteArr(T Begin, T End) {
  const char Kg = ' ', Edl = '\n';
  while (Begin != End) {
    qwrite(*Begin);
    ++Begin;
    putc_(Kg);
  }
  putc_(Edl);
}
struct Flusher_ {
  ~Flusher_() { flush(); }
} io_flusher;
}  // namespace IO
using namespace IO;
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long My_Rand(long long Mod) { return (unsigned long long)(rnd()) % Mod; }
template <class T1, class T2>
T1 Min(const T1 x, const T2 y) {
  return x > y ? y : x;
}
template <class T1, class T2>
T1 Max(const T1 x, const T2 y) {
  return x > y ? x : y;
}
template <class T1, class T2>
void To_max(T1 &x, const T2 y) {
  x < y ? x = y : x = x;
}
template <class T1, class T2>
void To_min(T1 &x, const T2 y) {
  x > y ? x = y : x = x;
}
inline long long qmul(long long x, long long y, long long p) {
  long long l = x * (y >> 25) % p * (1 << 25) % p;
  long long r = x * (y & ((1 << 25) - 1)) % p;
  return (l + r) % p;
}
inline long long qpow(long long n, long long base, long long mod = 1e18) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * base % mod;
    base = base * base % mod;
    n >>= 1;
  }
  return ret % mod;
}
inline long long gcd(long long x, long long y) {
  return !y ? x : gcd(y, x % y);
}
inline long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
void InitHash(unsigned long long &base, int &prime, unsigned long long &mod) {
  base = 131, prime = 233317, mod = 212370440130137957ll;
}
const int N = 4e6 + 5;
int n, p;
int f[N], q[N];
int main() {
  qread(n, p);
  for (int i = (1); i <= (n); ++i) {
    (f[i] += f[i - 1]) %= p;
    q[i] = f[i];
    if (i == 1) q[i] = 1;
    (f[i + 1] += q[i]) %= p;
    for (int j = (2); j <= (n / i); ++j) {
      (f[i * j] += q[i]) %= p;
      if (i * j + j <= n) (f[i * j + j] += p - q[i]) %= p;
    }
  }
  cout << (q[n] + p) % p << endl;
  return 0;
}
