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 <stdio.h>
#include <stdlib.h>

int N = 0;
int Q = 0;
int S1[100000];
int S2[100000];
int S3[100000];
int Com = 0;
int V1 = 0;
int V2 = 0;
int V3 = 0;
void relate();
void query();
void merge();

int main(){
	int i = 0;
	scanf("%d %d",&N,&Q);
	for(i = 0; i<N ; i++){
		S1[i] = i;
		S2[i] = 0;
		S3[i] = 1;
	}

	while(Q>0){
		scanf("%d",&Com);
		if(Com){
			scanf("%d %d",&V1, &V2);
			query();
		}else{
			scanf("%d %d %d",&V1 ,&V2, &V3);
			relate();
		}
		Q--;
	}
}

void relate(){
	//printf("R : %d %d %d\n", V1, V2, V3);
	if(S1[V1] == S1[V2]){
		return;
	}else{
		merge();
	}
}

void merge(){
	int i = 0;
	int s3 = S2[V2] - V3 -S2[V1];
	int s1 = S1[V1];
	int s2 = S1[V2];
	if(S3[s1] == 1){
		S3[s1]--;
		S3[s2]++;
		S1[V1] = s2;
		S2[V1] = S2[V1] + s3;
		//printf("Fast1\n");
		return;
	}else{ 
		if(S3[s2] == 1){
			S3[s1]++;
			S3[s2]--;
			S1[V2] = s1;
			S2[V2] = S2[V2]-s3;
			//printf("Fast2\n");
			return;
		}
	
	}

	for(i = 0; i<N ;i++){
		if(S1[i] == s1){
			//System.out.println("V1 = "+ V1 + "V2 = " + V2+ "V3 = " + V3);
			//printf("V1 = %d, V2 = %d, V3 = %d\n",V1, V2, V3);
			S3[s1]--;
			S3[s2]++;
			S1[i] = s2;
			S2[i] = S2[i] + s3;
			//printf("S2[%d] = %d\n", i, S2[i]);
			//System.out.println("S2[i] = "+ S2[i]);
		}else{
			continue;
		}
	}

}

void query(){
	//printf("Q : %d %d \n", V1, V2);
	if(S1[V1] == S1[V2]){
		printf("%d\n",S2[V2]-S2[V1]);
	}else{
		printf("?\n");
	}

}