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 <iostream>
#include <algorithm>
#include <queue>
#include <climits>
using namespace std;

int* pa;
int* v;
int n;
int q;
queue<int> qr;
void init(int n) {
	pa = new int[n];
	v = new int[n];
	for (int i = 0; i < n; ++i) {
		pa[i] = i;
		v[i] = 0;
	}
}

int find(int x) {
	if (x != pa[x]) {
		int p = pa[x];
		pa[x] = find(pa[x]);
		v[x] += v[p];
	}
	return pa[x];
}

void unite(int x, int y, int z) {
	int px = find(x);
	int py = find(y);
	if (px != py) {
		pa[px] = py;
		v[px] = v[y] - v[x] + z;
	}
}

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

int main() {
	cin >> n >> q;
	init(n);
	for(int i = 0; i < q; i++)
	{
		int op;
		cin >> op;
		if (op == 0) {
			int x, y, z;
			cin >> x >> y >> z;
			unite(x, y, z);
		}
		else {
			int x, y;
			cin >> x >> y;
			if (check(x, y)) {
				qr.push(v[x] - v[y]);
			}
			else {
				qr.push(INT_MAX);
			}
		}
	}
	while (!qr.empty())
	{
		int t = qr.front();
		if (t == INT_MAX)
			cout << "?" << endl;
		else
			cout << t << endl;
		qr.pop();
	}
	return 0;
}