CodeForces - 438D： The Child and Sequence（势能线段树）

At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:

1. Print operation l, r. Picks should write down the value of .
2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r).
3. Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x).

Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m (1 ≤ n, m ≤ 10⁵). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 10⁹) — initial value of array elements.

Each of the next m lines begins with a number type .

• If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1.
• If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 10⁹), which correspond the operation 2.
• If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 10⁹), which correspond the operation 3.

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Examples

Input

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3

Output

8
5

Input

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

Output

49
15
23
1
9

Note

Consider the first testcase:

• At first, a = {1, 2, 3, 4, 5}.
• After operation 1, a = {1, 2, 3, 0, 1}.
• After operation 2, a = {1, 2, 5, 0, 1}.
• At operation 3, 2 + 5 + 0 + 1 = 8.
• After operation 4, a = {1, 2, 2, 0, 1}.
• At operation 5, 1 + 2 + 2 = 5.

题意：给出数组，有三种操作，分别是区间求和，区间取模 ，单点修改。

思路：一个点被取模，那么其大小减半，所以一个数最多被操作log次，这样的话就不难想到势能线段树。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=100010;
int Mx[maxn<<2]; ll sum[maxn<<2];
void build(int Now,int L,int R)
{
    if(L==R){
        scanf("%d",&Mx[Now]);
        sum[Now]=Mx[Now]; return ;
    }
    int Mid=(L+R)>>1;
    build(Now<<1,L,Mid); build(Now<<1|1,Mid+1,R);
    Mx[Now]=max(Mx[Now<<1],Mx[Now<<1|1]);
    sum[Now]=sum[Now<<1]+sum[Now<<1|1];
}
ll query(int Now,int L,int R,int l,int r){
    if(l<=L&&r>=R) return sum[Now];
    int Mid=(L+R)>>1; ll res=0;
    if(l<=Mid) res+=query(Now<<1,L,Mid,l,r);
    if(r>Mid) res+=query(Now<<1|1,Mid+1,R,l,r);
    return res;
}
void change(int Now,int L,int R,int pos,int val)
{
    if(L==R){
        Mx[Now]=val; sum[Now]=val; return ;
    }
    int Mid=(L+R)>>1;
    if(pos<=Mid) change(Now<<1,L,Mid,pos,val);
    else change(Now<<1|1,Mid+1,R,pos,val);
    Mx[Now]=max(Mx[Now<<1],Mx[Now<<1|1]);
    sum[Now]=sum[Now<<1]+sum[Now<<1|1];
}
void modp(int Now,int L,int R,int l,int r,int P)
{
    if(Mx[Now]<P) return ;
    if(L==R) {
        Mx[Now]%=P; sum[Now]=Mx[Now]; return ;
    }
    int Mid=(L+R)>>1;
    if(l<=Mid) modp(Now<<1,L,Mid,l,r,P);
    if(r>Mid) modp(Now<<1|1,Mid+1,R,l,r,P);
    Mx[Now]=max(Mx[Now<<1],Mx[Now<<1|1]);
    sum[Now]=sum[Now<<1]+sum[Now<<1|1];
}
int main()
{
    int N,M,opt,L,R,P;
    scanf("%d%d",&N,&M);
    build(1,1,N);
    while(M--){
        scanf("%d",&opt);
        if(opt==1) {
            scanf("%d%d",&L,&R);
            printf("%I64d\n",query(1,1,N,L,R));
        }
        else if(opt==2){
            scanf("%d%d%d",&L,&R,&P);
            modp(1,1,N,L,R,P);
        }
        else {
            scanf("%d%d",&L,&R);
            change(1,1,N,L,R);
        }
    }
    return 0;
}