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:
// Weighted Union Find Trees

#include <iostream>

using namespace std;

int n, q;

int fat[100000];
int dis[100000];

int find(int u) {
	if (fat[u] == u)
		return u;
	int g = find(fat[u]);
	dis[u] += dis[fat[u]];
	fat[u] = g;
	return g;
}

void relate(int x, int y, int z) {
	int u = find(x);
	int v = find(y);
	fat[u] = v;
	dis[u] = z - dis[x] + dis[y];
}

bool check(int x, int y) {
	x = find(x);
	y = find(y);
	return x == y;
}

int diff(int x, int y) {
	find(x);
	find(y);
	return dis[x] - dis[y];
}

int main() {
	cin >> n >> q;
	for (int i = 0; i < n; ++i)
		fat[i] = i, dis[i] = 0;
	for (int i = 0; i < q; ++i) {
		int k, x, y, z;
		cin >> k >> x >> y;
		if (k == 0)
			cin >> z, relate(x, y, z);
		else {
			if (check(x, y))
				cout << diff(x, y) << endl;
			else
				cout << "?" << endl;
		}
	}
	return 0;
}