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 PYTHON3):

Solution:
n,q=map(int,input().split())
Q=[list(map(int,input().split())) for i in range(q)]

#UnionFind

Group=[[i,0] for i in range(n)]#グループ分け.Group[i]=jのときiとjは同じグループ

def find(x):#find(a)=find(b)のとき同じグループ
    ANS=Group[x][1]
    while Group[x][0] != x:
        x=Group[x][0]
        ANS+=Group[x][1]
    return x,ANS

def Union(x,y,z):  # xとyが同じグループになるよう更新
    if find(x)[0] != find(y)[0]:
        MIN=min(find(x)[0],find(y)[0])
        if find(x)[0]==MIN:
            Group[find(y)[0]]=[MIN,z+find(x)[1]-find(y)[1]]
        else:
            Group[find(x)[0]]=[MIN,-z+find(y)[1]-find(x)[1]]
            

for query in Q:
    if query[0]==0:
        Union(query[1],query[2],query[3])
        #print(Group)
    else:
        if find(query[1])[0]==find(query[2])[0]:
            print(find(query[2])[1]-find(query[1])[1])
        else:
            print("?")