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 n, q;
int p[100000], d[100000];

int find(int x) {
	if (p[x] == x)
		return x;
	else {
		int xp = p[x];
		p[x] = find(xp);
		d[x] += d[xp];
		return p[x];
	}
}

void relate(int x, int y, int z) {
	int xp = find(x), yp = find(y);

	if (xp != yp) {
		p[yp] = xp;
		d[yp] = d[x] + z - d[y];
	}
}

int main()
{
	cin >> n >> q;

	for (int i = 0; i < n; i++) {
		p[i] = i;
		d[i] = 0;
	}

	for (int i = 0; i < q; i++) {
		int c, x, y, z;

		cin >> c;

		if (c == 0) {
			cin >> x >> y >> z;
			relate(x, y, z);
		}
		else {
			cin >> x >> y;
			if (find(x) == find(y))
				cout << d[y] - d[x] << endl;
			else
				cout << "?" << endl;
		}
	}
}