【Codeforces Round 362 (Div 2)C】【STL-map 最近公共祖先思想】Lorenzo Von Matterhorn 数域二叉树的路径权值变更查询

C. Lorenzo Von Matterhorn

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Barney lives in NYC. NYC has infinite number of intersections numbered with positive integers starting from 1. There exists a bidirectional road between intersections i and 2i and another road between i and 2i + 1 for every positive integer i. 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 will qconsecutive events happen soon. There are two types of events:

1. Government makes a new rule. A rule can be denoted by integers v, u and w. As the result of this action, the passing fee of all roads on the shortest path from u to v increases by w dollars.

2. Barney starts moving from some intersection v and goes to intersection u 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 integer q (1 ≤ q ≤ 1 000).

The next q lines contain the information about the events in chronological order. Each event is described in form 1 v u w if it's an event when government makes a new rule about increasing the passing fee of all roads on the shortest path from u to v by w dollars, or in form 2 v u if it's an event when Barnie goes to cuddle from the intersection v to the intersection u.

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

Note

In the example testcase:

Here are the intersections used:

1. Intersections on the path are 3, 1, 2 and 4.
2. Intersections on the path are 4, 2 and 1.
3. Intersections on the path are only 3 and 6.
4. Intersections on the path are 4, 2, 1 and 3. Passing fee of roads on the path are 32, 32and 30 in order. So answer equals to 32 + 32 + 30 = 94.
5. Intersections on the path are 6, 3 and 1.
6. Intersections on the path are 3 and 7. Passing fee of the road between them is 0.
7. Intersections on the path are 2 and 4. Passing fee of the road between them is 32(increased by 30 in the first event and by 2 in the second).

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;int n;map<LL, LL>mop;void add(LL x, LL y, LL z){while (x != y){if (y > x)swap(x, y);mop[x] += z;x >>= 1;}}LL check(LL x, LL y){LL sum = 0;while (x != y){if (y > x)swap(x, y);if (mop.count(x))sum += mop[x];x >>= 1;}return sum;}int main(){while (~scanf("%d", &n)){mop.clear();int op;LL x, y, z;for (int i = 1; i <= n; ++i){scanf("%d", &op);if (op == 1){scanf("%lld%lld%lld", &x, &y, &z);add(x, y, z);}else{scanf("%lld%lld", &x, &y);printf("%lld\n", check(x, y));}}}return 0;}/*【题意】对于正整数域上的二叉树，x的左儿子为2x，右儿子为2x+1有q（1000）次操作。对于每次操作，有两种类型：1 x,y,z 我们增加从x到y路径上每条边的权值为z2 x y 询问从x到y的路径边权【类型】STL - map 最近公共祖先思想【分析】虽然数域范围很大，但是操作数很小，我们可以用一个mapmop[x]表示x到其父节点的路径上的权值累计量。然后我们就可以O(lognlogn)的复杂度维护每次询问啊【时间复杂度&&优化】O(qlognlogn)*/