Description

You have N integers, A₁, A₂, ... ,A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁,A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9
15

Hint

題意很簡單，區間增減，區間求和。

學習fhq treap，每一次區間操作把區間cut出來，操作完在merge回去，很神奇，debug兩天，終於ac啦。

程式碼：

/* ***********************************************
Author :xianxingwuguan
Created Time :2014/3/5 22:03:29
File Name :1.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
ll ran(){
	static ll ranx=323232;
	return ranx+=ranx<<2|1;
}
struct Node{
	Node *l,*r;
	ll delta,sum,size,fix,key;
}*root,*nill,mem[1101000];
ll tot;
void up(Node *o){
	if(o==nill)return;
	o->size=o->l->size+o->r->size+1;
	o->sum=o->key+o->delta;
	if(o->l!=nill)o->sum+=o->l->sum;
	if(o->r!=nill)o->sum+=o->r->sum;
}
void down(Node *o){
	if(o->delta){
		if(o->l!=nill)o->l->delta+=o->delta,o->l->sum+=o->l->size*o->delta,o->l->key+=o->delta;
		if(o->r!=nill)o->r->delta+=o->delta,o->r->sum+=o->r->size*o->delta,o->r->key+=o->delta;
		o->delta=0;
	}
}
void newnode(Node *&o,ll val){
	o=&mem[tot++];
	o->l=o->r=nill;
	o->size=1;
	o->fix=ran();
	o->sum=o->key=val;
	o->delta=0;
}
void cut(Node *o,Node *&a,Node *&b,int p) {
        if (o->size <= p) a = o,b = nill;
        else if (p==0) a = nill,b = o;
        else {
			down(o);
                if (o->l->size >= p){
                     b = o;
                     cut(o->l,a,b->l,p);
                     up(b);
                } 
				else{
                    a = o;
                    cut(o->r,a->r,b,p - o->l->size - 1);
                    up(a);
                }
        }
}
void merge(Node *&o,Node *a,Node *b){
    if (a==nill) o = b;
    else if (b==nill) o = a;
    else{
        if (a->fix> b->fix){
		     down(a);
             o = a;
             merge(o->r,a->r,b);
        } 
		else{
			down(b);
            o = b;
            merge(o->l,a,b->l);
        }
		up(o);
    }
}
void add(ll l,ll r,ll val){
	Node *a,*b,*c;
	cut(root,a,b,l-1);
	cut(b,b,c,r-l+1);
	b->key+=val;
	b->delta+=val;
	b->sum+=val*b->size;
	merge(a,a,b);
	merge(root,a,c);
}
ll query(ll l,ll r){
	Node *a,*b,*c;
	cut(root,a,b,l-1);
	cut(b,b,c,r-l+1);
	ll ans=b->sum;
	merge(b,b,c);
	merge(root,a,b);
	return ans;
}
void del(ll p) {
     Node *a,*b,*c;
     cut(root,a,b,p-1);
     cut(b,b,c,1);
     merge(root,a,c);
}
Node *Ins(ll x){
	if(x==0)return nill;
	if(x==1){
		int d;
		Node *a;
		scanf("%lld",&d);
		newnode(a,d);
		return a;
	}
	Node *a,*b;
	int m=x>>1;
	a=Ins(m);
	b=Ins(x-m);
	merge(a,a,b);
	return a;
}
void ins(ll pos,ll x){
     ll d;
	 Node *a,*b,*c,*t;
	 cut(root,a,b,pos);
	 c=Ins(x);
	 merge(a,a,c);
	 merge(root,a,b);
}
int main()
{
     //freopen("data.in","r",stdin);
    // freopen("data.out","w",stdout);
     ll i,j,k,m,n;
	 while(~scanf("%lld%lld",&n,&m)){
		 tot=0;
		 newnode(nill,0);
		 nill->size=0;
		 root=nill;
		 ins(0,n);
		 while(m--){
			 char op[3];
			 scanf("%s",op);
			 if(op[0]=='Q'){
				 ll l,r;
				 scanf("%lld%lld",&l,&r);
				 printf("%lld\n",query(l,r));
			 }
			 else {
				 scanf("%lld%lld%lld",&i,&j,&k);
				 add(i,j,k);
			 }
		 }
		 /*Node *a,*b,*c;
		 cut(root,a,b,0);
		 cut(b,b,c,1);
		 cout<<a->size<<" "<<b->size<<" "<<c->size<<endl;
		 cout<<b->key<<" "<<b->delta<<" "<<b->sum<<endl;
		 */
	 }
     return 0;
}
/*
10 15
7 2 3 4 5 6 7 8 9 10
C 1 1 2
C 1 1 10000
C 2 2 1000
C 1 3 3
C 1 4 4
C 1 5 5
C 1 6 3
C 1 7 7
C 1 8 8
C 1 9 9
C 1 10 10
Q 1 1
Q 1 2
Q 1 10
Q 5 10

*/