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:
#line 1 "test/datastructure/potentialunionfind.test.cpp"
#define PROBLEM "http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_1_B"
#define CXX "g++"

#include <bits/stdc++.h>
using namespace std;
using lint = long long;

#line 1 "test/datastructure/../../library/datastructure/unionfindpotential.cpp"
template <class Abel>
struct UnionFindPotential {
    vector<int> parent, rank;
    vector<Abel> diff_weight;

    UnionFindPotential(int n = 1, Abel SUM_UNITY = 0)
        : parent(n + 1), rank(n + 1, 0), diff_weight(n + 1, SUM_UNITY) {
        iota(parent.begin(), parent.end(), 0);
    }
    int root(int x) {
        if (parent[x] == x) {
            return x;
        } else {
            int r = root(parent[x]);
            diff_weight[x] += diff_weight[parent[x]];
            return parent[x] = r;
        }
    }
    Abel weight(int x) {
        root(x);
        return diff_weight[x];
    }
    bool isSame(int x, int y) { return root(x) == root(y); }
    bool unite(int x, int y, Abel w) {
        w += weight(x);
        w -= weight(y);
        x = root(x);
        y = root(y);
        if (x == y)
            return false;
        if (rank[x] < rank[y])
            swap(x, y), w = -w;
        if (rank[x] == rank[y])
            ++rank[x];
        parent[y] = x;
        diff_weight[y] = w;
        return true;
    }
    Abel diff(int x, int y) { return weight(y) - weight(x); }
};
#line 9 "test/datastructure/potentialunionfind.test.cpp"

int main() {
    std::cin.tie(0);
    std::ios_base::sync_with_stdio(false);
    int n, q;
    cin >> n >> q;
    UnionFindPotential<lint> uf(n);
    for (int i = 0; i < q; ++i) {
        int t;
        cin >> t;
        if (t == 0) {
            int x, y, z;
            cin >> x >> y >> z;
            uf.unite(x, y, z);
        } else {
            int x, y;
            cin >> x >> y;
            if (uf.isSame(x, y))
                cout << uf.diff(x, y) << "\n";
            else
                cout << "?"
                     << "\n";
        }
    }
    return 0;
}