coderforce Educational Codeforces Round 6 E.New Year Tree

E. New Year Tree

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 vertex1.

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 c_i (1 ≤ c_i ≤ 60) — the colour of thei-th vertex.

Each of the next n - 1 lines contains two integersx_j, y_j (1 ≤ x_j, y_j ≤ n) — the vertices of thej-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 integert_k (1 ≤ t_k ≤ 2) — the type of thek-th query. For the queries of the first type then follows two integersv_k, c_k (1 ≤ v_k ≤ n, 1 ≤ c_k ≤ 60) — the number of the vertex whose subtree will be recoloured with the colour c_k. For the queries of the second type then follows integerv_k (1 ≤ v_k ≤ 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

7 101 1 1 1 1 1 11 21 31 43 53 63 71 3 22 11 4 32 11 2 52 11 6 42 12 22 3

Output

234512

Input

23 301 2 2 6 5 3 2 1 1 1 2 4 5 3 4 4 3 3 3 3 3 4 61 21 31 42 52 63 73 84 94 104 116 126 137 147 157 168 178 1810 1910 2010 2111 2211 232 12 52 62 72 82 92 102 112 41 12 11 13 11 14 11 15 11 16 11 17 11 18 11 19 11 20 11 21 11 22 11 23 12 12 52 62 72 82 92 102 112 4

Output

613321235512211123

给你一棵树每次更新或者查询子树的颜色种类，或者更新某一棵子树上面的颜色为某一种，颜色总数为60，那么一看就类似与区间更新用一个longlong的来保存该数里面的颜色，也就是状压，但是这是树上面的操作，可以用DFS把一颗树转为一个长线段，L和R数字表示以该节点为根的子树的区间范围，那么就转换成了一个普通的区间更新，最后注意细节和long long的维护。

#include<bits/stdc++.h>//#include<windows.h>using namespace std;#define maxn 400050int n,m,tid;int c[maxn];int L[maxn],R[maxn];int ID[maxn];vector<int>mp[maxn];struct node{    int l;    int r;    long long flag;    long long cur;}tree[maxn*8];void build(int id,int l,int r){    tree[id].l=l;    tree[id].r=r;    tree[id].flag=0;    if(l==r)    {        tree[id].cur=(long long)((long long)1<<c[ID[l]]);        return ;    }    int mid=(l+r)/2;    build(id*2,l,mid);    build(id*2+1,mid+1,r);    tree[id].cur=tree[id*2].cur|tree[id*2+1].cur;}void pushup(int id){    if(tree[id].flag!=0)    {        tree[id*2].flag=tree[id].flag;        tree[id*2+1].flag=tree[id].flag;        tree[id*2].cur=tree[id*2].flag;        tree[id*2+1].cur=tree[id*2+1].flag;        tree[id].flag=0;    }}void update(int id,int x,int y,long long z){    int l=tree[id].l;    int r=tree[id].r;    if(r<x||l>y)        return ;    if(l>=x&&r<=y)    {        pushup(id);        tree[id].flag=z;        tree[id].cur=z;        return ;    }    pushup(id);    update(id*2,x,y,z);    update(id*2+1,x,y,z);    tree[id].cur=tree[id*2].cur|tree[id*2+1].cur;}long long query(int id,int x,int y){    int l=tree[id].l;    int r=tree[id].r;    int mid=(l+r)/2;    if(r<x||l>y)        return 0;    if(l>=x&&r<=y)    {        pushup(id);        return tree[id].cur;    }    pushup(id);    return query(id*2,x,y)|query(id*2+1,x,y);}int getsum(long long cur){    long long tmp=cur;    int sum=0;    while(tmp)    {        sum+=tmp%2;        tmp/=2;    }    return sum;}void dfs(int now){    if(L[now]!=0)        return ;    L[now]=++tid;    int len=mp[now].size();    for(int i=0;i<len;i++)        dfs(mp[now][i]);    R[now]=tid;}int main(){    scanf("%d%d",&n,&m);    tid=0;    memset(L,0,sizeof(L));    for(int i=1;i<=n;i++)        scanf("%d",&c[i]);    for(int i=1;i<=n-1;i++)    {        int x,y;        scanf("%d%d",&x,&y);        mp[x].push_back(y);        mp[y].push_back(x);    }    dfs(1);    for(int i=1;i<=n;i++)        ID[L[i]]=i;    build(1,1,n);    while(m--)    {        int t;        scanf("%d",&t);        if(t==1)        {            int x,y;            scanf("%d%d",&x,&y);            update(1,L[x],R[x],(long long)1<<y);        }        else        {            int z;            scanf("%d",&z);            long long ans=query(1,L[z],R[z]);            printf("%d\n",getsum(ans));        }    }return 0;}