/*

51 nod 1421 最大MOD值

problem:

从数组中选择两个数使 a[i]%a[j]最大

solve:

可以发现a[i]越接近a[j]的倍数则越大. 所以枚举a[j]的倍数然后二分查找.

最开始以为这个会T....

M+M/2+...M=O(MlgM)

hhh-2016/09/16-16:33:10

*/

#pragma comment(linker,"/STACK:124000000,124000000")

#include <algorithm>

#include <iostream>

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <vector>

#include <math.h>

#include <queue>

#include <set>

#include <map>

#define lson  i<<1

#define rson  i<<1|1

#define ll long long

#define clr(a,b) memset(a,b,sizeof(a))

#define scanfi(a) scanf("%d",&a)

#define scanfs(a) scanf("%s",a)

#define scanfl(a) scanf("%I64d",&a)

#define scanfd(a) scanf("%lf",&a)

#define key_val ch[ch[root][1]][0]

#define eps 1e-7

#define inf 0x3f3f3f3f3f3f3f3f

using namespace std;

const ll mod = 1e9+7;

const int maxn = 200010;

const double PI = acos(-1.0);

template<class T> void read(T&num)

{

    char CH;

    bool F=false;

    for(CH=getchar(); CH<'0'||CH>'9'; F= CH=='-',CH=getchar());

    for(num=0; CH>='0'&&CH<='9'; num=num*10+CH-'0',CH=getchar());

    F && (num=-num);

}

int stk[70], tp;

template<class T> inline void print(T p)

{

    if(!p)

    {

        puts("0");

        return;

    }

    while(p) stk[++ tp] = p%10, p/=10;

    while(tp) putchar(stk[tp--] + '0');

    putchar('\n');

}

int a[maxn];

int main()

{

    int n;

    read(n);

    for(int i =0; i < n; i++)

    {

        read(a[i]);

    }

    sort(a,a+n);

//    int cnt = unique(a,a+n)-a;

    int ans = 0;

    for(int i =0; i < n; i++)

    {

        if(a[i] == 1)

            continue;

        for(int j = 1;j <= a[n-1]/a[i] + 1; j++)

        {

            int pos = lower_bound(a+i,a+n,j * a[i]) - a;

            pos -- ;

            if(pos > 0 && pos <= n && a[pos] > a[i])

                ans = max(ans, a[pos] % a[i]);

            if(ans == a[i]-1)

                break;

//            cout << j <<" " << pos <<" " <<ans << endl;

        }

    }

    print(ans);

    return 0;

}