You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... ,Ab.

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

The sums may exceed the range of 32-bit integers.

题目大意：对于一个区间有两种操作：

1.Q即访问，访问某个区间，求这个区间的最大值。

2.C即插入，在指定的区间每个元素都加一个值

思路：线段树最基本的操作，改变一个区间的值，求一个区间的和。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
long long sum[500000],lazy[500000];
void creat(int l,int r,int o)//建立线段树
{
    if(l==r)
    {
        scanf("%lld",&sum[o]);
        return;
    }
    int mid=(l+r)>>1;
    creat(l,mid,o<<1);
    creat(mid+1,r,o<<1|1);
    sum[o]=sum[o<<1]+sum[o<<1|1];
}
void push(int o,int l)//更新左儿子和右儿子，下移操作
{
    if(lazy[o])
    {
        lazy[o<<1]+=lazy[o];
        lazy[o<<1|1]+=lazy[o];
        sum[o<<1]+=lazy[o]*(l-(l>>1));
        sum[o<<1|1]+=lazy[o]*(l>>1);
        lazy[o]=0;//**********
    }
}
void update(int x,int y,int l,int r,int o,int c)
{
    if(l>=x&&r<=y)//****x和y为所要的区间。l-r为访问的区间
    {
        sum[o]+=c*(r-l+1);
        lazy[o]+=c;
        return;
    }
    push(o,r-l+1);
    int mid=(r+l)>>1;
    if(mid>=x)
        update(x,y,l,mid,o<<1,c);
    if(y>mid)
        update(x,y,mid+1,r,o<<1|1,c);
    sum[o]=sum[o<<1]+sum[o<<1|1];
}
long long ask(int x,int y,int l,int r,int o)
{
    if(l>=x&&r<=y)
        return sum[o];
    push(o,r-l+1);
    int mid=(r+l)>>1;
    long long num=0;
    if(mid>=x)
        num+=ask(x,y,l,mid,o<<1);
    if(y>mid)
        num+=ask(x,y,mid+1,r,o<<1|1);
    return num;
}
int main()
{
    int n,q;
    while(~scanf("%d%d",&n,&q))
    {
        memset(sum,0,sizeof(sum));
        memset(lazy,0,sizeof(lazy));
        creat(1,n,1);
        while(q--)
        {
            char a[10];
            int l,r,c;
            scanf("%s",a);
            if(a[0]=='Q')
            {
                scanf("%d%d",&l,&r);
                printf("%lld\n",ask(l,r,1,n,1));
            }
            else
            {
                scanf("%d%d%d",&l,&r,&c);
                update(l,r,1,n,1,c);
            }
        }
    }
    return 0;
}