Problem: There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions.

* relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$
* diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$

Constraints

* $2 \leq n \leq 100,000$
* $1 \leq q \leq 200,000$
* $0 \leq x, y < n$
* $x \ne y$
* $0 \leq z \leq 10000$
* There are no inconsistency in the given information

Input


$n \; q$
$query_1$
$query_2$
:
$query_q$


In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format.


0 $x \; y\; z$


or


1 $x \; y$


where '0' of the first digit denotes the relate information and '1' denotes the diff question.

Output

For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$.

Example

Input

5 6
0 0 2 5
0 1 2 3
1 0 1
1 1 3
0 1 4 8
1 0 4


Output

2
?
10

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
using ll = long long;

int Par[100001];
ll diff_weight[100001];


void init(int n)
{
	for(int i = 0;i<n;i++)
	{
		Par[i] = i;
	}
}

int find(int x)
{
	if(Par[x]==x)return x;
	int f = find(Par[x]);
	diff_weight[x] += diff_weight[Par[x]];//親全部足す
	return Par[x] = f;
}

int weight(int x)
{
	find(x);
	return diff_weight[x];
}

int diff(int x,int y)
{
	if(find(x)!=find(y))return INT_MIN;
	return weight(y)-weight(x);
}

void unite(int x,int y,int w)//yはxよりw大きくする。
{
	w += weight(x); w-=weight(y);
	x = find(x); y = find(y);
	if(x==y)return;
	Par[y]=x;
	diff_weight[y] = w;
}



int main()
{
	int n,q;
	cin >> n >> q;
	//cin >> n >> q;
	init(n+1);
	while(q--)
	{
		int op;
		cin >> op;
		if(op)
		{
			int x,y;
			cin >> x >> y;
			if(diff(x,y)!=INT_MIN)cout<<diff(x,y)<<'\n';
			else cout<<'?'<<endl;
		}
		else
		{
			int x,y,w;
			cin >> x >> y >> w;
			unite(x,y,w);
		}
		
	}
}