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 <stdio.h>
#include <stack>
using namespace std;
const int N=1e5+10;
int p[N],val[N];
stack<int>s[N];
int find(int n){
    if(p[n]==n)return n;
    return p[n]=find(p[n]);
}
int main(){
    int n,q,l,r,temp,a,b;
    scanf("%d%d",&n,&q);
    for(int i=0;i<n;i++){
        p[i]=i;
        val[i]=0;
        s[i].push(i);
    }
    while(q--){
        scanf("%d%d%d",&temp,&l,&r);
        if(temp==1){
            if(find(l)!=find(r))printf("?\n");
            else printf("%d\n",val[r]-val[l]);
        }
        else{
            scanf("%d",&temp);
            a=find(l);
            b=find(r);
            if(a==b)continue;
            if(s[a].size()>s[b].size()){
                p[b]=a;
                temp+=val[l]-val[r];
                while(!s[b].empty()){
                    val[s[b].top()]+=temp;
                    s[a].push(s[b].top());
                    s[b].pop();
                }
            }
            else{
                p[a]=b;
                temp=val[r]-temp-val[l];
                while(!s[a].empty()){
                    val[s[a].top()]+=temp;
                    s[b].push(s[a].top());
                    s[a].pop();
                }
            }
        }
    }
    
}