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.

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, A_a+1, ... , A_b.

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

 #include "cstdio" //poj 3468 lazy操作

 #include "cstring"

 #include "iostream"

 using namespace std;

 #define N 100005

 #define LL long long

 struct node{

     int x,y;

     LL sum;

     LL add;   //记录以当前节点为根节点的树中需要增加的值

 }a[*N];

 void Build(int t,int x,int y)

 {

     a[t].x = x;

     a[t].y = y;

     a[t].sum = a[t].add = ;

     if(a[t].x == a[t].y)  //到了叶子节点

     {

         scanf("%lld",&a[t].sum);

         return ;

     }

     int mid = (a[t].x + a[t].y)/;

     Build(t<<,x,mid);

     Build(t<<|,mid+,y);

     a[t].sum = a[t<<].sum + a[t<<|].sum;

 }

 void Push_down(int t)  //将add(增值)向下推一级

 {

     LL add = a[t].add;

     a[t<<].add += add;

     a[t<<|].add += add;

     a[t<<].sum += add*(a[t<<].y-a[t<<].x+);

     a[t<<|].sum += add*(a[t<<|].y-a[t<<|].x+);

     a[t].add = ;

 }

 LL Query(int t,int x,int y)

 {

     if(a[t].x==x &&a[t].y==y)

         return a[t].sum;

     Push_down(t);

     int mid = (a[t].x + a[t].y)/;

     if(y<=mid)

         return Query(t<<,x,y);

     if(x>mid)

         return Query(t<<|,x,y);

     else

         return Query(t<<,x,mid) + Query(t<<|,mid+,y);

 }

 void Add(int t,int x,int y,int k)

 {

     if(a[t].x==x && a[t].y==y) //不在推到叶子节点，t下的子孙要增加的值存入a[t].add中

     {

         a[t].add += k;

         a[t].sum += (a[t].y-a[t].x+)*k;

         return ;

     }

     a[t].sum += (y-x+)*k;

     Push_down(t);

     int mid = (a[t].x+a[t].y)/;

     if(y<=mid)

         Add(t<<,x,y,k);

     else if(x>mid)

         Add(t<<|,x,y,k);

     else

     {

         Add(t<<,x,mid,k);

         Add(t<<|,mid+,y,k);

     }

 }

 int main()

 {

     int n,m;

     char ch;

     int x,y,k;

     scanf("%d %d",&n,&m);

     Build(,,n);

     while(m--)

     {

         getchar();

         scanf("%c %d %d",&ch,&x,&y);

         if(ch=='Q')

             printf("%lld\n",Query(,x,y));

         else

         {

             scanf("%d",&k);

             Add(,x,y,k);

         }

     }

     return ;

 }