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 arr[100000], value[100000];

int find(int x){
	if (arr[x] == x)return x;
	else {
		int fx = arr[x];
		arr[x] = find(arr[x]);
		value[x] += value[fx];
		return arr[x];
	}
	return arr[x] == x ? x : arr[x] = find(arr[x]);
}

void relate(int a, int b, int c)
{
	int fa = find(a), fb = find(b);
	if (fa != fb){
		value[fa] = c + value[b] - value[a];
		arr[fa] = fb;
	}
}

int main()
{
	int n, q, ch, a, b, c;
	scanf("%d%d", &n, &q);
	for (int i = 0; i < n; ++i)arr[i] = i;
	
	while (q--){
		scanf("%d", &ch);
		if (ch){
			scanf("%d%d", &a, &b);
			if (find(a) == find(b))cout << value[a] - value[b] << endl;
			else printf("?\n");
		}else{
			scanf("%d%d%d", &a, &b, &c);
			relate(a, b, c);
		}
	}
	return 0;
}