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<vector>
#include<algorithm>
#include<utility>
#include<string>
#include<cmath>
#include<cstring>
#include<queue>
#include<map>
#include<climits>
#include<queue>

#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define llong long long
#define pb(a) push_back(a)
using namespace std;
typedef pair<int, int> P;
typedef long long int ll;
typedef pair<ll,ll> LLP;
int dx[4] = {1,0,0,-1};
int dy[4] = {0,1,-1,0};
#define INF 1000000000007
#define MOD 1000000007
#define ARRAY_MAX 3005


//重み付きUnion-Find
struct Potential_UnionFind{

    vector<int> par;
    vector<int> rank;
    vector<int> diff_weight;

    void init(int n) {
        par.resize(n);
        rank.resize(n);
        diff_weight.resize(n);
        REP(i,n){
            par[i] = i;
            rank[i] = 0;
            diff_weight[i] = 0;
        }
    }

    int root(int x) {//親を探す
        if (par[x] == x) {
            return x;
        }
        else {
            int r = root(par[x]);
            diff_weight[x] += diff_weight[par[x]];
            return par[x] = r;
        }
    }

    int weight(int x) {
        root(x);
        return diff_weight[x];
    }

    bool issame(int x,int y) {
        return root(x) == root(y);
    }

    void merge(int x, int y,int w) {

        int rx = root(x);
        int ry = root(y);
        
        if(rank[rx] < rank[ry]){
            diff_weight[rx] = -w - weight(x) + weight(y);
            par[rx] = ry;
        }else{
            diff_weight[ry] = w + weight(x) - weight(y);
            par[ry] = rx;
            if(rank[rx] == rank[ry]){
                rank[rx]++;
            }
        }
    }

    int diff(int x, int y) {
        return weight(y) - weight(x);
    }
};


int main(){

    Potential_UnionFind tree;
    int n,q;
    cin >> n >> q;
    tree.init(n);
    for(int i = 0;i < q;i++){

        int a,b,c,d;
        cin >> a;
        if(a == 0){
            cin >> b >> c >> d;
            tree.merge(b,c,d);
        }else{
            cin >> b >> c;
            if(tree.issame(b,c)){
                cout << tree.diff(b,c) << endl;
            }else{
                cout << "?" << endl;
            }
        }


    }


    return 0;
}