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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 andQ. 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.

Output

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

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

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

题意：给你n个整数，有两种操作，第一种，当字母为Q时，输出区间【a,b】的和，为C时，将区间【a,b】整体加c;

思路：线段树模板，注意数据范围诶，另外延迟标记用的可以^__^,哈哈哈；

下面附上代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;const int M=100005;struct Tree{LL sum;LL lazy;LL l,r,len; };Tree tree[M<<2];void pushup(LL o){tree[o].sum=tree[ll].sum+tree[rr].sum;}void pushdown(LL o){if(tree[o].lazy){tree[ll].lazy+=tree[o].lazy;tree[rr].lazy+=tree[o].lazy;tree[ll].sum+=tree[o].lazy*tree[ll].len;tree[rr].sum+=tree[o].lazy*tree[rr].len;tree[o].lazy=0;}}void build(LL o, LL L,LL R){tree[o].l=L,tree[o].r=R;tree[o].len=R-L+1;tree[o].lazy=0;if(L==R){LL p;scanf("%lld",&p);tree[o].sum=p;return ;}LL mid=(L+R)>>1;build(o<<1,L,mid);build(o<<1|1,mid+1,R);pushup(o);}void update(LL o,LL L,LL R,LL v){if(tree[o].l>=L&&tree[o].r<=R){tree[o].lazy+=v;tree[o].sum+=v*tree[o].len;   return ;   }pushdown(o);LL mid=(tree[o].l+tree[o].r)>>1;if(R<=mid)update(ll,L,R,v);else if(L>mid)update(rr,L,R,v);else{update(ll,L,mid,v);update(rr,mid+1,R,v);}pushup(o);}LL Query(LL o,LL L,LL R){if(tree[o].l>=L&&tree[o].r<=R)return tree[o].sum;LL mid=(tree[o].l+tree[o].r)>>1;pushdown(o);if(R<=mid)return Query(ll,L,R);else if(L>mid)return Query(rr,L,R);elsereturn Query(ll,L,mid)+Query(rr,mid+1,R);}int main(){LL N,Q1;char s[3]={0};LL a,b,c;scanf("%lld %lld",&N,&Q1);build(1,1,N);while(Q1--){scanf("%s",s);if(s[0]=='Q'){scanf("%lld %lld",&a,&b);LL k=Query(1,a,b);printf("%lld\n",k);}else{scanf("%lld %lld %lld",&a,&b,&c);update(1,a,b,c);}}return 0;}