CodeForces346 C. Number Transformation II

C. Number Transformation II

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a sequence of positive integers x1, x2, ..., xn and two non-negative integers a and b. Your task is to transform a into b. To do that, you can perform the following moves:

• subtract 1 from the current a;
• subtract a mod xi (1 ≤ i ≤ n) from the current a.

Operation a mod xi means taking the remainder after division of number a by number xi.

Now you want to know the minimum number of moves needed to transform a into b.

Input

The first line contains a single integer n (1 ≤  n ≤ 105). The second line contains n space-separated integers x1, x2, ..., xn(2 ≤  xi ≤ 109). The third line contains two integers a and b (0  ≤ b ≤  a ≤ 109, a - b ≤ 106).

Output

Print a single integer — the required minimum number of moves needed to transform number a into number b.

Examples

input

33 4 530 17

output

6

input

35 6 71000 200

output

206

/*CodeForces346  C. Number Transformation II给你两个数a,b1.每次对于当前的a减去12.每次对于当前的a减去 a%(ta[i])求最少多少次能得到b类似于贪心,每次减去尽可能多的值。但是一直TLE- -.   后来可以发现a~a-a%ta[i]的所有值如果减去a%ta[i],都会等于同一个值。  如果当a-a%ta[i] < b时,ta[i]可以说在后面的搜索就没有作用了.于是乎把ta[i]除去.hhh-2016-04-16 17:15:20*/#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <map>#include <algorithm>#include <functional>#include <math.h>using namespace std;#define lson  (i<<1)#define rson  ((i<<1)|1)typedef long long ll;const int mod = 1000000009;const int maxn = 100040;int ta[maxn];int main(){    int n;    while( scanf("%d",&n) != EOF)    {        int num = 0;        int a,b;        for(int i =0; i < n; i++)        {            scanf("%d",&ta[i]);        }        sort(ta,ta+n);        n = unique(ta,ta+n)-ta;        scanf("%d%d",&a,&b);        while(a > b)        {            int cur = a - 1;            for(int i =n-1; i >= 0; i--)            {                if(a-a%ta[i] >= b)                    cur = min(a-a%ta[i],cur);            }            a = cur;            num ++;            if(a == b) break;            while(n)            {                if(a-a%ta[n-1] < b)                    n--;                else                    break;            }        }        printf("%d\n",num);    }    return 0;}