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:
def root(x):
    if(par[x]==x):
        return x
    else:
        r=root(par[x])
        diff_weight[x]+=diff_weight[par[x]]
        par[x]=r
        return r

def weight(x):
    root(x)
    return diff_weight[x]

def diff(x,y):
    return weight(y)-weight(x)

def relate(a,b,w):
    w+=weight(a);w-=weight(b)
    a=root(a)
    b=root(b)
    if(a==b):
        return False
    if rank[a]<rank[b]:
        a,b=b,a
        w=-w
    if rank[a]==rank[b]:
        rank[b]+=1
        par[b]=a
    par[b]=a
    diff_weight[b]=w
    siz[a]+=siz[b]
    return True

def size(a):
    return siz[root(a)]
    
def same(a,b):
    return root(a)==root(b)
    
N,Q=map(int,input().split())
par=[i for i in range(N)]
siz=[1 for _ in range(N)]
rank=[0 for _ in range(N)]
diff_weight=[0 for _ in range(N)]
for i in range(Q):
    query=list(map(int,input().split()))
    if not query[0]:
        X,Y,W=query[1:]
        relate(X,Y,W)
    else:
        X,Y=query[1:]
        if same(X,Y):
            print(diff(X,Y))
        else:
            print('?')