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;
int f[1000005];
int v[1000005];
int find(int x){
	if(f[x]==x){
		return x;
	}else{
		int fx=f[x];
		f[x]=find(f[x]);
		v[x]+=v[fx];
		return f[x];
	}
//	return f[x]==x?x:f[x]=find(f[x]);
}
void join(int x,int y,int z){
	// x->y z
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy){
		v[fx]=z+v[y]-v[x];
		f[fx]=fy;
	}
}
int main(){
	int n,q,t,l,r;
	cin>>n>>q;
	for(int i=0;i<=n;i++)f[i]=i;
	while(q--){
		cin>>t>>l>>r;
		if(t){
			int fx=find(l);
			int fy=find(r);
			if(fx==fy){
				cout<<v[l]-v[r]<<endl;
			}else cout<<"?"<<endl;
		}else{
			int z;cin>>z;
			join(l,r,z);
		}
	}
	return 0;
}