Segment tree (B - Beza's Hangover)

Template question (query interval, to modify (with map to look at the wine store mL))

Friday nights are tricky for UFPE's ICPC competitors - they must be careful with their plans, after all, they train on saturdays and must be in good shape to help their team. Beza, however, went to a party on his friday night and said that no booze could bring him down. He was wrong.

Beza's night had NN hours, the party's bar had MM distinct beverages and, at each of the NN hours, Beza went to the bar and got one of the MM drinks. Lucas, who is an experienced drinker, said that a competitor would not be hungover the next day if the total volume of alcohol he drank did not exceed half the number of minutes he spent drinking.

Saturday morning, Beza woke up and noticed he was terribly hungover, and this made him think about the things he did on that night. Since Beza has got a nasty headache, he's not able to think clearly, and, therefore, asks you to help him answer some questions he has about that night.

Beza's questions can be of two types:

• 11 XX YY - he wonders if it would be a good idea to drink YY at the XX-th hour. Therefore, he changes the beverage of the XX-th hour to YY. It is guaranteed that the beverage YY was available at the bar. (1≤X≤N1≤X≤N)
• 22 LL RR - he asks himself if he would be hungover on saturday, if his night only consisted of drinking from the LL-th to the RR-th hour, given his night's "drink schedule" at the moment of this query. (1≤L≤R≤N1≤L≤R≤N)

Input

The first line of input consists of three integers NN, MM, and QQ, (1≤N,M,Q≤2⋅1051≤N,M,Q≤2⋅105). The next line contains NN strings DiDi, 1≤|Di|≤201≤|Di|≤20, each one describing the name of the ii-th drink Beza had that night. Then, MM lines follow, each one containing two entries SS and VV, 1≤|S|≤201≤|S|≤20, 1≤V≤1001≤V≤100, which describe, respectively, the name of a drink the bar had and how many liters of alcohol it had. Finally, QQ lines follow, each one containing three integers, as described on the problem's statement.

Output

For all queries of type 22, you must answer "YES" if Beza would be hungover the next day, given the interval of hours on which he would be drinking, and "NO" otherwise. It is guaranteed that there will be at least one query of type 22.

Example

Input

6 6 5 
vodka pie beats whiskey vodka cuba 
vodka 30 
caipirinha 10 
pie 35 
beats 15 
whiskey 20 
Cuba 50 
2 3 4 
1 3 cuba 
2 3 3 
1 5 cuba 
2 1 5

Output

NO
YES
YES

Note

The total amount of hours passed by on a given interval [L,R][L,R] is R−L+1R−L+1.

Meaning of the questions: given three numbers n, m, k, respectively, what wine to drink for several hours, representatives m m kinds of wine, the recording capacity of these wines, k this inquiry.

When asked whether these wines in this period of time and whether the person can be drunk (if the number is less than half milliliter drink drinking equal time not drunk, or drunk)

Code:

#include <bits/stdc++.h>

using namespace std;

const int maxn =2e5+100;
int a[maxn];
int tree[maxn*4];
int n,m,k;
map<string,int>mp;
char s[maxn][22];
void build(int l,int r,int node)
{
    if(l==r)
    {
        tree[node]=a[l];
        return ;
    }
    int mid=(l+r)/2;
    build(l,mid,node*2);
    build(mid+1,r,node*2+1);
    tree[node]=tree[node*2]+tree[node*2+1];
}
void updata(int node,int l,int r,int x,int y)
{
    if(l==r)
    {
        tree[node]=y;
        return ;
    }
    int mid=(l+r)/2;
    if(x<=mid)
    {
        updata(node*2,l,mid,x,y);
    }
    if(x>mid)
    {
        updata(node*2+1,mid+1,r,x,y);
    }
    tree[node]=tree[node*2]+tree[node*2+1];
}
int query(int node,int l,int r,int x,int y)
{
    if(l>=x&&r<=y)
    {
        return tree[node];
    }
    int mid=(l+r)/2;
    int res=0;
    if(x<=mid)
    {
        res+=query(node*2,l,mid,x,y);
    }
    if(y>mid)
    {
        res+=query(node*2+1,mid+1,r,x,y);
    }
    return res;
}
int main()
{
//    ios::sync_with_stdio(false);
//    cin.tie(0);
//    cout.tie(0);
    scanf("%d%d%d",&n,&m,&k);
    for(int i=0;i<n;i++)
    {
        scanf("%s",s[i]);
    }
    char str[maxn];
    int x;
    while(m--)
    {
        scanf("%s%d",str,&x);
        mp[str]=x;
//        cout<<mp[str]<<" "<<str<<endl;
    }
    for(int i=1;i<=n;i++)
    {
        a[i]=mp[s[i-1]];
//        cout<<s[i-1]<<" "<<mp[s[i-1]]<<endl;
//        cout<<a[i]<<" "<<s[i-1]<<endl;
    }
//    cout<<endl;
    build(1,n,1);
    int u,v,w;
    while(k--)
    {
        scanf("%d",&u);
        if(u==1)
        {
            scanf("%d",&v);
            scanf("%s",str);
            updata(1,1,n,v,mp[str]);
        }
        else
        {
            scanf("%d%d",&v,&w);
            int ans=query(1,1,n,v,w);
//            printf("%d\n",ans);
            if(ans>(w-v+1)*30) printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}