Problem: 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

Solution (in CPP):

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

template <typename T,typename E>
struct SegmentTree{
  typedef function<T(T,T)> F;
  typedef function<T(T,E)> G;
  typedef function<T(E,E)> H;
  typedef function<E(E,int)> P;
  int n;
  F f;
  G g;
  H h;
  P p;
  T d1;
  E d0;
  vector<T> dat;
  vector<E> laz;
  SegmentTree(){};
  SegmentTree(int n_,F f,G g,H h,T d1,E d0,
	      vector<T> v=vector<T>(),P p=[](E a,int b){return a;}):
    f(f),g(g),h(h),d1(d1),d0(d0),p(p){
    init(n_);
    if(n_==(int)v.size()) build(n_,v);
  }
  void init(int n_){
    n=1;
    while(n<n_) n*=2;
    dat.clear();
    dat.resize(2*n-1,d1);
    laz.clear();
    laz.resize(2*n-1,d0);
  }
  void build(int n_, vector<T> v){
    for(int i=0;i<n_;i++) dat[i+n-1]=v[i];
    for(int i=n-2;i>=0;i--)
      dat[i]=f(dat[i*2+1],dat[i*2+2]);
  }
  void update(int a,int b,E x,int k,int l,int r){
    if(r<=a||b<=l) return;
    if(a<=l&&r<=b){
      laz[k]=h(laz[k],x);
      return;
    }
    update(a,b,x,k*2+1,l,(l+r)/2);
    update(a,b,x,k*2+2,(l+r)/2,r);
    dat[k]=f(g(dat[k*2+1],p(laz[k*2+1],(r-l)>>1)),
	     g(dat[k*2+2],p(laz[k*2+2],(r-l)>>1)));
  }
  void update(int a,int b,E x){
    update(a,b,x,0,0,n);
  }
  T query(int a,int b,int k,int l,int r){
    if(r<=a||b<=l) return d1;
    if(a<=l&&r<=b) return g(dat[k],p(laz[k],r-l));
    T vl=query(a,b,k*2+1,l,(l+r)/2);
    T vr=query(a,b,k*2+2,(l+r)/2,r);
    return g(f(vl,vr),p(laz[k],min(r,b)-max(l,a)));
  }
  T query(int a,int b){
    return query(a,b,0,0,n);
  }
};


signed main(){
  int n,q;
  cin>>n>>q;
  SegmentTree<Int,Int> ch(n,
			  [](Int a,Int b){return a+b;},
			  [](Int a,Int b){return a+b;},
			  [](Int a,Int b){return a+b;},
			  0,0,vector<Int>(n,0),
			  [](Int a,int b){return a*b;});
  for(int i=0;i<q;i++){
    int c,s,t,x;
    cin>>c;
    if(c){
      cin>>s>>t;
      cout<<ch.query(s-1,t)<<endl;
    }else{
      cin>>s>>t>>x;
      ch.update(s-1,t,x);
    }
  }
}

/*
verified on 2017/10/15
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_2_G
*/