Problem:
Score : 400 points

Problem Statement

Given are a sequence A= {a_1,a_2,......a_N} of N positive even numbers, and an integer M.

Let a semi-common multiple of A be a positive integer X that satisfies the following condition for every k (1 \leq k \leq N):

There exists a non-negative integer p such that X= a_k \times (p+0.5).

Find the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Constraints

1 \leq N \leq 10^5

1 \leq M \leq 10^9

2 \leq a_i \leq 10^9

a_i is an even number.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
a_1 a_2 ... a_N

Output

Print the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Sample Input 1

2 50
6 10

Sample Output 1

2

15 = 6 \times 2.5

15 = 10 \times 1.5

45 = 6 \times 7.5

45 = 10 \times 4.5

Thus, 15 and 45 are semi-common multiples of A. There are no other semi-common multiples of A between 1 and 50, so the answer is 2.

Sample Input 2

3 100
14 22 40

Sample Output 2

0

The answer can be 0.

Sample Input 3

5 1000000000
6 6 2 6 2

Sample Output 3

166666667

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#include <vector>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define MOOD 1000000007

using namespace std;
using ll = long long;
using P = pair<int,int>;

int cmpnum(const void *a, const void *b)
{
    long long *A = (long long *)a;
    long long *B = (long long *)b;
    if (*A > *B)
        return 1;
    if (*A < *B)
        return -1;
    return 0;
}

int char_sort_max( const void * a , const void * b ) {
    return strcmp(( char * )a , ( char * )b );
}

int char_sort_min( const void * a , const void * b ) {
    return strcmp(( char * )a , ( char * )b );
}

int gcd(int x, int y)
{
    int t;

    while (y != 0) {
        t = x % y;  x = y;  y = t;
    }
    return x;
}
int lcm(int a, int b)
{
    return a * b / gcd(a, b);
}

struct UnionFind{
    vector<int> d;
    UnionFind(int n = 0):d(n,-1){};

    int find(int x){
        if(d[x] < 0) {
            return x;
        }
        else{
            return d[x] = find(d[x]);
        }
    }
    bool unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) return false;
        if (d[x] > d[y]) swap(x, y);

        d[x] += d[y];
        d[y] = x;
        return true;
    }
    bool same(int x, int y){
        return find(x) == find(y);
    }
    int size(int x){
        return -d[find(x)];
    }

};

struct mint {
    ll x;

    mint(ll x) : x((x % MOOD + MOOD) % MOOD) {}

    mint operator-() const { return mint(-x); }

    mint &operator+=(const mint a) {
        if ((x += a.x) >= MOOD) x -= MOOD;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += MOOD-a.x) >= MOOD) x -= MOOD;
        return *this;
    }

    mint &operator*=(const mint a) {
        if ((x *= a.x) %= MOOD);
        return *this;
    }

    mint operator+(const mint a)const {
        mint res(*this);
        return res += a;
    }

    mint operator-(const mint a) const {
        mint res(*this);
        return res -= a;
    }

    mint operator*(const mint a) const {
        mint res(*this);
        return res *= a;
    }

    mint pow(ll t) const{
        if(!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if(t&1) a *= *this;
        return a;
    }

    mint inv() const {
        return pow(MOOD-2);
    }

    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }

    mint operator/(const mint a) const {
        mint res(*this);
        return res/=a;
    }

};

mint kaijyo(int n) {
    if (n == 0) return 1;
    mint x = kaijyo(n/2);
    x *= x;
    if (n%2 == 1) x *= 2;
    return x;
}

mint choose(int n, int a) {
    mint x = 1, y = 1;
    rep(i,a) {
        x *= n-i;
        y *= i+1;
    }
    return x / y;
}


int shakutori(ll x, ll n,ll k, ll a[])
{
    long long int ai,res,ok,ng,mid;
    res = 0;
    for (int i = 0; i < n; i++)
    {
        // printf("%lld\n", res);
        ai = a[i];
        if(ai == 0){
            if(x < 0){
                res += 0;
            } else {
                res += n - 1;
            }
        } else
        {
            if(ai <= -1){
                if (ai * ai <= x){
                    res -= 1;
                }
                ok = n;
                ng = -1;
                while (ok-ng>1)
                {
                    mid = (ok + ng) / 2;
                    if(a[mid]*ai<=x){
                        ok = mid;
                    } else {
                        ng = mid;
                    }
                }
                res += n - ok;
            } else
            {
                if (ai * ai <= x)
                {
                    res -= 1;
                }
                ok = -1;
                ng = n;
                while (ng - ok > 1)
                {
                    mid = (ok + ng) / 2;
                    if (a[mid] * ai <= x)
                    {
                        ok = mid;
                    }
                    else
                    {
                        ng = mid;
                    }
                }
                res += ok + 1;
            }

        }
    }
    // printf("%lld\n", res);
    return k <= res / 2;
}

int two_n(int tei, int jyousuu){
    if(jyousuu == 0) return 1;
    int x = two_n(tei,jyousuu/2);
    x *= x ;
    if(jyousuu % 2 == 1) {
        x *= tei;
    }
    return x;




}


int main() {
    int n,m;
    cin >> n >> m;
    int a[n],b[n];
    int lcm_a,lcm_b;
    for(int i = 0; i < n; ++i){
        cin >> a[i];
        b[i] = a[i] * 0.5;
    }
    for(int i = 0; i < n-1; ++i){
         lcm_a = lcm(a[i],a[i+1]);
        lcm_b = lcm(b[i],b[i+1]);
    }
    printf("%d",m/lcm_b -m/lcm_a);



}

int A(){
    int A,X;
    scanf("%d %d",&A,&X);
    if(A*500 >= X){
        printf("Yes");
    }
    else printf("No");
    return 0;
}

int B(){
    char S[100];
    int A;
    scanf("%d",&A);
    scanf("%s",S);
    int X = strlen(S);
    int count = 0;
    for(int i = 0; i <= X - 3; ++i){
        if(S[i]=='A' && S[i+1]=='B' && S[i+2] =='C'){
            count++;
        }
    }
    printf("%d",count);
}

int C(){
    int n;
    cin >> n;
    vector<int> x (n);
    vector<int> p(n);
    vector<int> q(n);
    for(int i = 0; i < n; ++i) {
        cin >> p[i];
    }
    for(int i = 0; i < n; ++i) {
        cin >> q[i];
    }
    for(int i = 0; i < n; ++i)x[i]= i+1;
    vector<vector<int>> A;
    int q_count,p_count;

    do{
        A.push_back(x);
        if(x == p){
            p_count = A.size();
        }
        if(x == q){
            q_count = A.size();
        }
    }while(next_permutation(x.begin(),x.end()));

    printf("%d",abs(p_count-q_count));

    return 0;
}