Problem: Igor K. very much likes a multiplayer role playing game WineAge II. Who knows, perhaps, that might be the reason for his poor performance at the university. As any person who plays the game, he is interested in equipping his hero with as good weapon and outfit as possible. 

One day, as he was reading the game's forum yet again, he discovered a very interesting fact. As it turns out, each weapon in the game is characterised with k different numbers: a1, ..., ak. They are called hit indicators and according to the game developers' plan they are pairwise coprime. 

The damage that is inflicted during a hit depends not only on the weapon's characteristics, but also on the hero's strength parameter. Thus, if the hero's strength equals n, than the inflicted damage will be calculated as the number of numbers on the segment <image>, that aren't divisible by any hit indicator ai.

Recently, having fulfilled another quest, Igor K. found a new Lostborn sword. He wants to know how much damage he will inflict upon his enemies if he uses it.

Input

The first line contains two integers: n and k (1 ≤ n ≤ 1013, 1 ≤ k ≤ 100). They are the indicator of Igor K's hero's strength and the number of hit indicators.

The next line contains space-separated k integers ai (1 ≤ ai ≤ 1000). They are Lostborn sword's hit indicators. The given k numbers are pairwise coprime.

Output

Print the single number — the damage that will be inflicted by Igor K.'s hero when he uses his new weapon. 

Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.

Examples

Input

20 3
2 3 5


Output

6


Input

50 2
15 8


Output

41

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long getint() {
  long long _x = 0, _tmp = 1;
  char _tc = getchar();
  while ((_tc < '0' || _tc > '9') && _tc != '-') _tc = getchar();
  if (_tc == '-') _tc = getchar(), _tmp = -1;
  while (_tc >= '0' && _tc <= '9') _x *= 10, _x += (_tc - '0'), _tc = getchar();
  return _x * _tmp;
}
inline long long add(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x + _y;
  if (_ >= _mod) _ -= _mod;
  return _;
}
inline long long sub(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x - _y;
  if (_ < 0) _ += _mod;
  return _;
}
inline long long mul(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x * _y;
  if (_ >= _mod) _ %= _mod;
  return _;
}
long long mypow(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 1ll;
  long long _tmp = mypow(_a, _x / 2, _mod);
  _tmp = mul(_tmp, _tmp, _mod);
  if (_x & 1) _tmp = mul(_tmp, _a, _mod);
  return _tmp;
}
long long mymul(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 0ll;
  long long _tmp = mymul(_a, _x / 2, _mod);
  _tmp = add(_tmp, _tmp, _mod);
  if (_x & 1) _tmp = add(_tmp, _a, _mod);
  return _tmp;
}
inline bool equal(double _x, double _y) {
  return _x > _y - 1e-9 && _x < _y + 1e-9;
}
int __ = 1, _cs;
long long n;
int k;
vector<int> a;
void build() {}
void init() {
  n = getint();
  k = getint();
  for (int i = 0; i < k; i++) a.push_back(getint());
  sort((a).begin(), (a).end());
  reverse((a).begin(), (a).end());
}
int dp[50000][101];
bool got[50000][101];
long long DP(int now, long long res) {
  if (res == 0 || now == k) return res;
  if (res < 50000 && dp[res][now] != -1) return dp[res][now];
  long long vl = DP(now + 1, res) - DP(now + 1, res / a[now]);
  if (res < 50000) dp[res][now] = vl;
  return vl;
}
void solve() {
  memset(dp, -1, sizeof(dp));
  printf("%lld\n", DP(0, n));
}
int main() {
  build();
  while (__--) {
    init();
    solve();
  }
}