#include <iostream>
#define int long long
using namespace std;

int n,q;
int all[300000],part[300000];

void init(){
	int nn = 1;
	while(nn < n) nn *= 2;
	n = nn;
}

void add(int a,int b,int x,int k,int l,int r){
	if(b <= l || r <= a) return;
	if(a <= l && r <= b){
		all[k] += x;
		return;
	}
	part[k] += (min(b,r) - max(a,l)) * x;
	add(a,b,x,k * 2 + 1,l,(l + r) / 2);
	add(a,b,x,k * 2 + 2,(l + r) / 2,r);
}

int get(int a,int b,int k,int l,int r){
	int res = 0;
	if(a <= l && r <= b) return all[k] * (r - l) + part[k];
	if(b <= l || r <= a) return 0;
	res += all[k] * (min(b,r) - max(a,l));
	res += get(a,b,k * 2 + 1,l,(l + r) / 2);
	res += get(a,b,k * 2 + 2,(l + r) / 2,r);
	return res;
}

signed main(){
	cin >> n >> q;
	init();
	for(int i = 0;i < q;i++){
		int v;
		cin >> v;
		if(v == 0){
			int s,t,x;
			cin >> s >> t >> x;
			add(s - 1,t,x,0,0,n);
		}else{
			int s,t;
			cin >> s >> t;
			cout << get(s - 1,t,0,0,n) << endl;
		}
	}
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:

* add(s, t, x): add x to as, as+1, ..., at.
* getSum(s, t): report the sum of as, as+1, ..., at.



Note that the initial values of ai (i = 1, 2, . . . , n) are 0.

Constraints

* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 1 ≤ s ≤ t ≤ n
* 0 ≤ x < 1000

Input


n q
query1
query2
:
queryq


In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:


0 s t x


or


1 s t


The first digit represents the type of the query. '0' denotes add(s, t, x) and '1' denotes getSum(s, t).

Output

For each getSum operation, print the sum;

Examples

Input

3 5
0 1 2 1
0 2 3 2
0 3 3 3
1 1 2
1 2 3


Output

4
8


Input

4 3
1 1 4
0 1 4 1
1 1 4


Output

0
4

### Response

```cpp
#include <iostream>
#define int long long
using namespace std;

int n,q;
int all[300000],part[300000];

void init(){
	int nn = 1;
	while(nn < n) nn *= 2;
	n = nn;
}

void add(int a,int b,int x,int k,int l,int r){
	if(b <= l || r <= a) return;
	if(a <= l && r <= b){
		all[k] += x;
		return;
	}
	part[k] += (min(b,r) - max(a,l)) * x;
	add(a,b,x,k * 2 + 1,l,(l + r) / 2);
	add(a,b,x,k * 2 + 2,(l + r) / 2,r);
}

int get(int a,int b,int k,int l,int r){
	int res = 0;
	if(a <= l && r <= b) return all[k] * (r - l) + part[k];
	if(b <= l || r <= a) return 0;
	res += all[k] * (min(b,r) - max(a,l));
	res += get(a,b,k * 2 + 1,l,(l + r) / 2);
	res += get(a,b,k * 2 + 2,(l + r) / 2,r);
	return res;
}

signed main(){
	cin >> n >> q;
	init();
	for(int i = 0;i < q;i++){
		int v;
		cin >> v;
		if(v == 0){
			int s,t,x;
			cin >> s >> t >> x;
			add(s - 1,t,x,0,0,n);
		}else{
			int s,t;
			cin >> s >> t;
			cout << get(s - 1,t,0,0,n) << endl;
		}
	}
	return 0;
}
```