codeforces 697C Lorenzo Von Matterhorn（二叉树LCA）

Lorenzo Von Matterhorn

Barney lives in NYC. NYC has infinite number of intersections numbered with positive integers starting from 1. There exists a bidirectional road between intersectionsi and2i and another road betweeni and2i + 1 for every positive integeri. You can clearly see that there exists a unique shortest path between any two intersections.

Initially anyone can pass any road for free. But since SlapsGiving is ahead of us, there willq consecutive events happen soon. There are two types of events:

1. Government makes a new rule. A rule can be denoted by integersv,u andw. As the result of this action, the passing fee of all roads on the shortest path fromu tov increases byw dollars.

2. Barney starts moving from some intersectionv and goes to intersectionu where there's a girl he wants to cuddle (using his fake name Lorenzo Von Matterhorn). He always uses the shortest path (visiting minimum number of intersections or roads) between two intersections.

Government needs your calculations. For each time Barney goes to cuddle a girl, you need to tell the government how much money he should pay (sum of passing fee of all roads he passes).

Input

The first line of input contains a single integerq (1 ≤ q ≤ 1 000).

The next q lines contain the information about the events in chronological order. Each event is described in form 1vuw if it's an event when government makes a new rule about increasing the passing fee of all roads on the shortest path fromu tov byw dollars, or in form2vu if it's an event when Barnie goes to cuddle from the intersectionv to the intersectionu.

1 ≤ v, u ≤ 10¹⁸, v ≠ u, 1 ≤ w ≤ 10⁹ states for every description line.

Output

For each event of second type print the sum of passing fee of all roads Barney passes in this event, in one line. Print the answers in chronological order of corresponding events.

Example

Input

71 3 4 301 4 1 21 3 6 82 4 31 6 1 402 3 72 2 4

Output

94032

题目大意：给出一个二叉树，有两种操作，第一种：把两个点u，v之间所有的边都 +w ； 第二种求出u，v两点之间的距离。

解题思路：因为是二叉树所以在这道题中任意两个节点之间的最多有128条边，所以可以采用比较暴力的做法来增加或计算两点之间的距离，方法是这样的，每次从两点中选取较大的那个点，将他除以二，也就是找到他的祖先，这样若干次操作一定能找到他们两个的最近公共祖先。

AC代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include<set>#include<map>#include<queue>#define maxn 1005#define inf 0x3f3f3f3f#define ONES(x) __builtin_popcount(x)using namespace std;typedef long long ll ;const double eps =1e-8;const int mod = 1000000007;map<ll,ll>road;int main(){    //freopen("test.txt","r",stdin);    ll q,rule,u,v,w,ans;    cin>>q;    while(q--)    {        cin>>rule;        if(rule==1)        {            cin>>u>>v>>w;            while(u!=v)            {                if(u<v)                    swap(u,v);//make sure that u > v                road[u] += w;                //cout << u << ' ' << road[u] << endl;                //road[v] += w;                u /= 2;            }        }        else        {            cin>>u>>v;            ans = 0;            while(u!=v)            {                if(u<v)                    swap(u,v);//make sure that u > v                ans += road[u];                //ans += road[v];                u /= 2;            }            cout << ans << endl;        }    }    //cout << "Hello world!" << endl;    return 0;}