CodeForces - 1101D：GCD Counting （樹分治）

You are given a tree consisting of n vertices. A number is written on each vertex; the number on vertex i is equal to ai

.

Let's denote the function g(x,y)

as the greatest common divisor of the numbers written on the vertices belonging to the simple path from vertex x to vertex y (including these two vertices). Also let's denote dist(x,y) as the number of vertices on the simple path between vertices x and y, including the endpoints. dist(x,x)=1 for every vertex x

.

Your task is calculate the maximum value of dist(x,y)

among such pairs of vertices that g(x,y)>1

.

Input

The first line contains one integer n

— the number of vertices (1≤n≤2⋅105)

.

The second line contains n

integers a1, a2, ..., an (1≤ai≤2⋅105)

— the numbers written on vertices.

Then n−1

lines follow, each containing two integers x and y (1≤x,y≤n,x≠y) denoting an edge connecting vertex x with vertex y

. It is guaranteed that these edges form a tree.

Output

If there is no pair of vertices x,y

such that g(x,y)>1, print 0. Otherwise print the maximum value of dist(x,y)

among such pairs.

Examples

Input

3
2 3 4
1 2
2 3

Output

1

Input

3
2 3 4
1 3
2 3

Output

2

Input

3
1 1 1
1 2
2 3

Output

0

題意：讓你求最長的路徑長度，滿足路上gcd不為1；

思路：分治的做法比較暴力，但是時限比較長，有板子就直接上了。 由於gcd具有收斂性，路徑上的gcd不會太多，而且越遠回越接近1，我們記錄每一個gcd的最深的位置即可。  （我用的以前的板子，所以用了map，此題的數據量可以不用map

還有一個思路，我們枚舉素因子，然后把含有這個素因子的點標記出來，求他們的最遠距離，更新答案。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=200010;
const int inf=0x7FFFFFFF;
int Laxt[maxn],Next[maxn<<1],To[maxn<<1],cnt,N,sn;
int a[maxn],sz[maxn],son[maxn],vis[maxn],root,res;
int dep[maxn];
ll ans[maxn];
map<int,int>mp,tp;
map<int,int>::iterator it1,it2;
inline void read(int &x) {
    x=0; char c=getchar();
    while(c>'9'||c<'0') c=getchar();
    while(c<='9'&&c>='0') x=(x<<3)+(x<<1)+c-'0',c=getchar();
}
void add(int u,int v){
    Next[++cnt]=Laxt[u];
    Laxt[u]=cnt; To[cnt]=v;
}
void getroot(int u,int fa) //找重心
{
    sz[u]=1; son[u]=0;
    for(int i=Laxt[u];i;i=Next[i]){
        if(To[i]!=fa&&!vis[To[i]]){
            getroot(To[i],u);
            sz[u]+=sz[To[i]];
            son[u]=max(son[u],sz[To[i]]);
        }
    }
    son[u]=max(son[u],sn-son[u]);
    if(root==0||son[root]>son[u]) root=u;
}
void getans(int u,int fa,int num) //對於當前鏈產生的新GCD
{
    dep[u]=dep[fa]+1; tp[num]=max(tp[num],dep[u]);
    for(int i=Laxt[u];i;i=Next[i]){
        if(!vis[To[i]]&&To[i]!=fa){
            getans(To[i],u,__gcd(num,a[To[i]]));
        }
    }
}
void solve(int u) //解決以u為根的子問題
{
    mp.clear(); mp[a[u]]=1; ans[a[u]]++; dep[u]=1;
    for(int i=Laxt[u];i;i=Next[i])
      if(!vis[To[i]]) {
         tp.clear(); getans(To[i],u,__gcd(a[u],a[To[i]]));
         for(it1=mp.begin();it1!=mp.end();it1++)
           for(it2=tp.begin();it2!=tp.end();it2++){
              int g=__gcd((*it1).first,(*it2).first);
              if(g>1) res=max(res,(*it1).second+(*it2).second-1);
        }
        for(it2=tp.begin();it2!=tp.end();it2++)
          mp[(*it2).first]=max((*it2).second,mp[(*it2).first]);
    }
}
void dfs(int u)  //分治
{
    vis[u]=1;  solve(u);
    for(int i=Laxt[u];i;i=Next[i]){
        if(vis[To[i]]) continue;
        root=0; sn=sz[To[i]];
        getroot(To[i],0); dfs(root);
    }
}
int main()
{
    read(N); int u,v,Max=0;
    for(int i=1;i<=N;i++) read(a[i]),Max=max(Max,a[i]);
    for(int i=1;i<N;i++) {
        read(u);read(v);
        add(u,v);  add(v,u);
    }
    if(Max>1) res=1;
    root=0; sn=N; getroot(1,0); dfs(root);
    printf("%d\n",res);
    return 0;
}