C. Vasya and Beautiful Arrays----思维题

C. Vasya and Beautiful Arrays

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Vasya's got a birthday coming up and his mom decided to give him an array of positive integers a of length n.

Vasya thinks that an array's beauty is the greatest common divisor of all its elements. His mom, of course, wants to give him as beautiful an array as possible (with largest possible beauty). Unfortunately, the shop has only one array a left.
On the plus side, the seller said that he could decrease some numbers in the array (no more than by k for each number).

The seller can obtain array b from array a if
the following conditions hold: bi > 0; 0 ≤ ai - bi ≤ k for
all 1 ≤ i ≤ n.

Help mom find the maximum possible beauty of the array she will give to Vasya (that seller can obtain).

Input

The first line contains two integers n and k (1 ≤ n ≤ 3·105; 1 ≤ k ≤ 106).
The second line contains n integers ai (1 ≤ ai ≤ 106) —
array a.

Output

In the single line print a single number — the maximum possible beauty of the resulting array.

Examples

input

6 1
3 6 10 12 13 16

output

3

input

5 38 21 52 15 77

output

7

Note

In the first sample we can obtain the array:

3 6 9 12 12 15

In the second sample we can obtain the next array:

7 21 49 14 77
题目链接：http://codeforces.com/contest/354/problem/C

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#define inf 0x3f3f3f3f
using namespace std;
int a[3000000];
int main(){
memset(a,0,sizeof(a));
int n,k;
scanf("%d%d",&n,&k);
int maxn=-1,minn=inf;
for(int i=0;i<n;i++){
int x;
scanf("%d",&x);
a[x]++;
maxn=max(x,maxn);
minn=min(x,minn);
}
for(int i=1;i<=maxn+k+1;i++){
a[i]+=a[i-1];
}
/*for(int i=1;i<=maxn;i++)
printf("%d ",a[i]);
printf("\n");*/
if(minn<=k+1)
printf("%d\n",minn);
else{
for(int i=minn;i>0;i--){
int ret=0;
for(int j=i;j<=maxn;j+=i){
ret+=a[j+k]-a[j-1];
}
if(ret==n){
printf("%d\n",i);
break;
}
}
}
return 0;
}