D. Recovering BST(dp+记忆化搜索)

Dima the hamster enjoys nibbling different things: cages, sticks, bad problemsetters and even trees!

Recently he found a binary search tree and instinctively nibbled all of its edges, hence messing up the vertices. Dima knows that if Andrew, who has been thoroughly assembling the tree for a long time, comes home and sees his creation demolished, he'll get extremely upset.

To not let that happen, Dima has to recover the binary search tree. Luckily, he noticed that any two vertices connected by a direct edge had their greatest common divisor value exceed 11.

Help Dima construct such a binary search tree or determine that it's impossible. The definition and properties of a binary search tree can be found here.

Input

The first line contains the number of vertices nn (2≤n≤7002≤n≤700).

The second line features nn distinct integers aiai (2≤ai≤1092≤ai≤109) — the values of vertices in ascending order.

Output

If it is possible to reassemble the binary search tree, such that the greatest common divisor of any two vertices connected by the edge is greater than 11, print "Yes" (quotes for clarity).

Otherwise, print "No" (quotes for clarity).

Examples

input

6
3 6 9 18 36 108

output

Yes

input

2
7 17

output

No

input

9
4 8 10 12 15 18 33 44 81

output

Yes

题目大概：

给出n个数，按升序给出，组成一颗二叉树，每两个相连节点之间必须有公约数。问是否能够组成这棵树。

思路：

预处理一下互质的数。

枚举根，dfs左右区间，dp+记忆化搜索看是否能找出一组解。

代码：

#include
#define ll long long
using namespace std;
const int maxn=800;
const int INF=1e9+7;
const int mod=1e9+7;
int a[maxn];
int zh[maxn][maxn];
int dp[maxn][maxn][2];
int vis[maxn][maxn][2];

int dfs(int l,int r,int k)
{
    if(l>r)return 1;
    if(vis[l][r][k])return dp[l][r][k];
    int flag;
    if(!k)flag=r+1;
    else flag=l-1;
    int flag1=0;
    for(int i=l;i<=r;i++)
    {
        if(!zh[i][flag]&&dfs(l,i-1,0)&&dfs(i+1,r,1))
        {
            flag1=1;
            break;
        }
    }
    vis[l][r][k]=1;
    return dp[l][r][k]=flag1;
}
long long gcd(long long m, long long n)
{
    while(m>0)
    {
        long long c = n % m;
        n = m;
        m = c;
    }
    return n;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=i;j<=n;j++)
        {
            if(gcd(a[i],a[j])==1)
            {
                zh[i][j]=zh[j][i]=1;
            }
        }
    }
    for(int i=1;i<=n;i++)
    {
        if(dfs(1,i-1,0)&&dfs(i+1,n,1))
        {
            printf("Yes\n");
            return 0;
        }
    }

    printf("No\n");
    return 0;
}