CodeForces 620E New Year Tree(线段树的骚操作第二弹)

The New Year holidays are over, but Resha doesn't want to throw away the New Year tree. He invited his best friends Kerim and Gural to help him to redecorate the New Year tree.

The New Year tree is an undirected tree with n vertices and root in the vertex 1.

You should process the queries of the two types:

1. Change the colours of all vertices in the subtree of the vertex v to the colour c.
2. Find the number of different colours in the subtree of the vertex v.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 4·105) — the number of vertices in the tree and the number of the queries.

The second line contains n integers ci (1 ≤ ci ≤ 60) — the colour of the i-th vertex.

Each of the next n - 1 lines contains two integers xj, yj (1 ≤ xj, yj ≤ n) — the vertices of the j-th edge. It is guaranteed that you are given correct undirected tree.

The last m lines contains the description of the queries. Each description starts with the integer tk (1 ≤ tk ≤ 2) — the type of the k-th query. For the queries of the first type then follows two integers vk, ck (1 ≤ vk ≤ n, 1 ≤ ck ≤ 60) — the number of the vertex whose subtree will be recoloured with the colour ck. For the queries of the second type then follows integer vk (1 ≤ vk ≤ n) — the number of the vertex for which subtree you should find the number of different colours.

Output

For each query of the second type print the integer a — the number of different colours in the subtree of the vertex given in the query.

Each of the numbers should be printed on a separate line in order of query appearing in the input.

Examples

input

7 10
1 1 1 1 1 1 1
1 2
1 3
1 4
3 5
3 6
3 7
1 3 2
2 1
1 4 3
2 1
1 2 5
2 1
1 6 4
2 1
2 2
2 3

output

2
3
4
5
1
2

input

23 30
1 2 2 6 5 3 2 1 1 1 2 4 5 3 4 4 3 3 3 3 3 4 6
1 2
1 3
1 4
2 5
2 6
3 7
3 8
4 9
4 10
4 11
6 12
6 13
7 14
7 15
7 16
8 17
8 18
10 19
10 20
10 21
11 22
11 23
2 1
2 5
2 6
2 7
2 8
2 9
2 10
2 11
2 4
1 12 1
1 13 1
1 14 1
1 15 1
1 16 1
1 17 1
1 18 1
1 19 1
1 20 1
1 21 1
1 22 1
1 23 1
2 1
2 5
2 6
2 7
2 8
2 9
2 10
2 11
2 4

output

6
1
3
3
2
1
2
3
5
5
1
2
2
1
1
1
2
3

题意：来自费老先生的翻译
你有一棵以1为根的有根树，有n个点，每个节点初始有一个颜色c[i]。

有两种操作：

1 v c 将以v为根的子树中所有点颜色更改为c

2 v 查询以v为根的子树中的节点有多少种不同的颜色

题解：来自蒟蒻xhk的题解：

我们看到这道题的颜色最多只有60种，所以理所应当的想到突破口就是颜色，60的范围可以考虑状压，其实我们在乎的只是有或者没有该颜色，所以我们可以用或运算来合并两个块，这自然是线段树的思路，

因为只有修改子树或者查询子树的操作，所以直接dfs序就可以啦~
对dfs序维护一个线段树，支持区间修改区间或，统计区间或之后得到的数在二进制下1的个数，这即为答案

代码如下：

#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define lson root<<1
#define rson root<<1|1
using namespace std;

struct node
{
    int l,r;
    long long sum,lazy;
} tr[1600010];

vector<int> g[400010];
int dfsn[400010],c[400010],w[400010],size[400010],tot;

int get_bit(long long x)
{
    int ans=0;
    while(x)
    {
        if(x&1)
        {
            ans++;
        }
        x>>=1;
    }
    return ans;
}

void push_up(int root)
{
    tr[root].sum=tr[lson].sum|tr[rson].sum;
}

void push_down(int root)
{
    tr[lson].sum=tr[root].lazy;
    tr[rson].sum=tr[root].lazy;
    tr[lson].lazy=tr[root].lazy;
    tr[rson].lazy=tr[root].lazy;
    tr[root].lazy=-1;
}

void build(int root,int l,int r)
{
    if(l==r)
    {
        tr[root].l=l;
        tr[root].r=r;
        tr[root].lazy=-1;
        tr[root].sum=1ll<<c[l];
        return ;
    }
    tr[root].l=l;
    tr[root].r=r;
    tr[root].lazy=-1;
    int mid=(l+r)>>1;
    build(lson,l,mid);
    build(rson,mid+1,r);
    push_up(root);
}

void update(int root,int l,int r,int val)
{
    if(tr[root].l==l&&tr[root].r==r)
    {
        tr[root].sum=1ll<<val;
        tr[root].lazy=1ll<<val;
        return ;
    }
    if(~tr[root].lazy)
    {
        push_down(root);
    }
    int mid=(tr[root].l+tr[root].r)>>1;
    if(l>mid)
    {
        update(rson,l,r,val);
    }
    else
    {
        if(r<=mid)
        {
            update(lson,l,r,val);
        }
        else
        {
            update(lson,l,mid,val);
            update(rson,mid+1,r,val);
        }
    }
    push_up(root);
}

long long query(int root,int l,int r)
{
    if(tr[root].l==l&&tr[root].r==r)
    {
        return tr[root].sum;
    }
    if(~tr[root].lazy)
    {
        push_down(root);
    }
    int mid=(tr[root].l+tr[root].r)>>1;
    if(l>mid)
    {
        return query(rson,l,r);
    }
    else
    {
        if(r<=mid)
        {
            return query(lson,l,r);
        }
        else
        {
            return query(lson,l,mid)|query(rson,mid+1,r);
        }
    }
}

void dfs(int now,int f)
{
    dfsn[now]=++tot;
    size[now]=1;
    c[tot]=w[now];
    for(int i=0;i<g[now].size();i++)
    {
        if(g[now][i]==f)
        {
            continue;
        }
        dfs(g[now][i],now);
        size[now]+=size[g[now][i]];
    }
}

int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&w[i]);
    }
    for(int i=1;i<n;i++)
    {
        int from,to;
        scanf("%d%d",&from,&to);
        g[from].push_back(to);
        g[to].push_back(from);
    }
    
    dfs(1,0);
    build(1,1,n);
    
    for(int i=1;i<=m;i++)
    {
        int kd;
        scanf("%d",&kd);
        if(kd==1)
        {
            int v,val;
            scanf("%d%d",&v,&val);
            update(1,dfsn[v],dfsn[v]+size[v]-1,val);
        }
        if(kd==2)
        {
            int v;
            scanf("%d",&v);
            printf("%d
",get_bit(query(1,dfsn[v],dfsn[v]+size[v]-1)));
        }
    }
}