題目連結：http://codeforces.com/problemset/problem/932/D

You are given a node of the tree with index 1 and with weight 0. Let cnt be the number of nodes in the tree at any instant (initially, cnt is set to 1). Support Q queries of following two types:

•  Add a new node (index cnt
   + 1) with weight W and add edge between node R and this node.
•  Output the maximum length of sequence of nodes which
  1. starts with R.
  2. Every node in the sequence is an ancestor of its predecessor.
  3. Sum of weight of nodes in sequence does not exceed X
     .
  4. For some nodes i, j that are consecutive in the sequence if i is an ancestor of j then w[i] ≥ w[j] and there should not exist a node k on simple path from i to j such that w[k] ≥ w[j]

The tree is rooted at node 1 at any instant.

Note that the queries are given in a modified way.

Input

First line containing the number of queries Q (1 ≤ Q ≤ 400000).

Let last be the answer for previous query of type 2 (initially last equals 0).

Each of the next Q lines contains a query of following form:

• 1 p q (1 ≤ p, q ≤ 1018): This is query of first type where  and . It is guaranteed that 1 ≤ R ≤ cnt and 0 ≤ W ≤ 109.
• 2 p q (1 ≤ p, q ≤ 1018): This is query of second type where  and . It is guaranteed that 1 ≤ R ≤ cntand 0 ≤ X ≤ 1015.

 denotes bitwise XOR of a and b.

It is guaranteed that at least one query of type 2 exists.

Output

Output the answer to each query of second type in separate line.

Examples

input

Copy

6
1 1 1
2 2 0
2 2 1
1 3 0
2 2 0
2 2 2

output

Copy

0
1
1
2

input

Copy

6
1 1 0
2 2 0
2 0 3
1 0 2
2 1 3
2 1 6

output

Copy

2
2
3
2

input

Copy

7
1 1 2
1 2 3
2 3 3
1 0 0
1 5 1
2 5 0
2 4 0

output

Copy

1
1
2

input

Copy

7
1 1 3
1 2 3
2 3 4
1 2 0
1 5 3
2 5 5
2 7 22

output

Copy

1
2
3

Note

In the first example,

last = 0

- Query 1: 1 1 1, Node 2 with weight 1 is added to node 1.

- Query 2: 2 2 0, No sequence of nodes starting at 2 has weight less than or equal to 0. last = 0

- Query 3: 2 2 1, Answer is 1 as sequence will be {2}. last = 1

- Query 4: 1 2 1, Node 3 with weight 1 is added to node 2.

- Query 5: 2 3 1, Answer is 1 as sequence will be {3}. Node 2 cannot be added as sum of weights cannot be greater than 1. last = 1

- Query 6: 2 3 3, Answer is 2 as sequence will be {3, 2}. last = 2

題目大意：

題目好繞啊！我斷斷續續讀了一天半，後來問dalao，才知道題目什麼意思，題讀懂了就是板子了，下面是題目大意：

給出一個樹，根節點是1，權值為0，然後又兩個操作：

1操作，在p節點下加一個權值為q的節點（注意異或last）

2操作，輸入p，q（對last異或）

從p節點開始（包括p節點），向上找祖先，符合要求的祖先：（權值>=子節點），並且符合要求的節點加起來不能超過輸入的那個q；

真是坑啊！一還是沒有注意到祖先是可以跳的（吐血！）

簡單的lca，還不用dfs建樹，就是要稍微改一下，還是比較簡單的（看我的註釋就知道改的多麼心力交瘁了。。。）

ac：

//#pragma comment(linker, "/STACK:1024000000,1024000000") 
 
#include<stdio.h>
#include<string.h>  
#include<math.h>  
#include<stdlib.h>

//#include<map>   
//#include<set>
#include<deque>  
#include<queue>  
#include<stack>  
#include<bitset> 
#include<string>  
#include<fstream>
#include<iostream>  
#include<algorithm>  
using namespace std;  
 
#define ll long long  
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b) 
#define clean(a,b) memset(a,b,sizeof(a))// 水印 
//std::ios::sync_with_stdio(false);
const int MAXN=1e6+10;
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;

//struct node{
//	ll v,nxt,w;
//	node(ll _v=0,ll _w=0,ll nxt=0):
//	v(_v),w(_w),nxt(_nxt){}
//}edge[MAXN<<1];
ll head[MAXN<<1],ecnt;
ll fa[MAXN][30],v[MAXN],sum[MAXN][30];
int n,m;

void intt()
{
	clean(head,-1);
	clean(fa,0);
	clean(v,0);
	clean(sum,0);
	ecnt=1;
}

void add(ll u,ll w)
{
	//cout<<"u下面放w："<<u<<" "<<w<<endl;
	ll temp=++ecnt;
	v[temp]=w;
	if(v[u]>=w)
		fa[temp][0]=u;
	else
	{
		for(int i=29;i>=0;--i)
		{
			if(v[fa[u][i]]<v[temp]&&fa[u][i])
				u=fa[u][i];
		}
		fa[temp][0]=fa[u][0];
	}
	//找到祖先 
	if(fa[temp][0])//存在祖先 
		sum[temp][0]=v[fa[temp][0]];//第一個祖先的權值即為向上求的sum 
	else
		sum[temp][0]=INF;
	//cout<<temp<<" :"<<fa[temp][0]<<endl;
	ll up=0,pre=fa[temp][0];
	while(fa[pre][up])//重新整理祖先 && 記錄該點~祖先的和 
	{
		
		fa[temp][up+1]=fa[pre][up];
		sum[temp][up+1]=sum[temp][up]+sum[fa[temp][up]][up];
		//cout<<temp<<": "<<(1<<(up+1))<<" :"<<fa[pre][up]<<endl;
		pre=fa[pre][up++];
	}
//	for(int i=0;i<30;++i)
//	{
//		cout<<fa[temp][i]<<" "<<sum[temp][i]<<endl;
//	}
//	cout<<endl;
}

ll lca(ll a,ll w)
{
//	for(int i=0;i<30;++i)
//		cout<<fa[a][i]<<" "<<sum[a][i]<<endl;
//	cout<<endl;
	if(v[a]>w)//本身>w 
		return 0;
	w=w-v[a];
	//cout<<w<<": "<<endl;
	ll ans=1;
	for(int i=29;i>=0;--i)
	{
		//cout<<i<<endl;
		//cout<<i<<": "<<w<<" "<<sum[a][i]<<" "<<fa[a][i]<<endl;
		if(w-sum[a][i]>=0&&fa[a][i])//祖先可以被重新整理到 && 祖先存在 
		{
			w=w-sum[a][i];//
			ans=ans+(1<<i);
			//cout<<w<<" :"<<ans<<" "<<fa[a][i]<<endl;
			a=fa[a][i];
			
		}
	}
	return ans;
}

int main()
{
	std::ios::sync_with_stdio(false);
    intt();
    ll last=0;
    cin>>n;
    while(n--)
    {
    	ll oper,p,q;
    	cin>>oper>>p>>q;
    	p=p^last,q=q^last;
    	if(oper==1)//插入q到p下面 
    		add(p,q);
		else//查詢從p開始所有<=q的節點 
		{
			last=lca(p,q);
			cout<<last<<endl;
		}
	}
}