codeforces 438D. The Child and Sequence(线段树)

D. The Child and Sequence

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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：区间求和 单点修改 区间取模

/*
标记没法维护
但可以发现一个性质，一个数每次取模最大也会比原来的1/2小
所以单点改，如果一个区间最大值都比模数小，就不用往后递归了。
*/
#include<iostream>
#include<cstdio>
#include<cstring>

#define N 100007
#define ll long long

using namespace std;
ll n,m,ans,cnt;
struct tree
{
    ll l,r,mx,sum;
}tr[N<<2];

inline ll read()
{
    ll x=0,f=1;char c=getchar();
    while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

void pushup(ll k)
{
    tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum;
    tr[k].mx=max(tr[k<<1].mx,tr[k<<1|1].mx);
}

void build(ll k,ll l,ll r)
{
    tr[k].l=l;tr[k].r=r;tr[k].sum=tr[k].mx=0;
    if(l==r)
    {
        tr[k].sum=read();
        tr[k].mx=tr[k].sum;
        return;
    }
    ll mid=l+r>>1;
    build(k<<1,l,mid);build(k<<1|1,mid+1,r);
    pushup(k);
}

void changemod(ll k,ll l,ll r,ll mod)
{
    if(l>r) return;
    if(tr[k].mx<mod) return;
    if(tr[k].l==tr[k].r)
    {
        tr[k].mx%=mod;
        tr[k].sum=tr[k].mx;
        return;
    }
    ll mid=tr[k].l+tr[k].r>>1;
    if(r<=mid) changemod(k<<1,l,r,mod);
    else if(l>mid) changemod(k<<1|1,l,r,mod);
    else changemod(k<<1,l,mid,mod),changemod(k<<1|1,mid+1,r,mod);
    pushup(k);
}

void change(ll k,ll pos,ll x)
{
    if(tr[k].l==tr[k].r && tr[k].l==pos)
    {
        tr[k].sum=tr[k].mx=x;
        return;
    }
    ll mid=tr[k].l+tr[k].r>>1;
    if(pos<=mid) change(k<<1,pos,x);
    if(pos>mid)  change(k<<1|1,pos,x);
    pushup(k);
}

ll query(ll k,ll l,ll r)
{
    if(l>r) return 0;
    if(tr[k].l==l && tr[k].r==r)
    return tr[k].sum;
    ll mid=tr[k].l+tr[k].r>>1;
    if(r<=mid) return query(k<<1,l,r);
    else if(l>mid) return query(k<<1|1,l,r);
    else return query(k<<1,l,mid)+query(k<<1|1,mid+1,r);
}

int main()
{
    n=read();m=read();
    build(1,1,n);ll opt,x,y,z;
    for(ll i=1;i<=m;i++)
    {
        opt=read();
        if(opt==1)
        {
            x=read();y=read();
            printf("%lld
",query(1,x,y));
        }
        if(opt==2)
        {
            x=read();y=read();z=read();
            changemod(1,x,y,z);
        }
        if(opt==3)
        {
            x=read();y=read();
            change(1,x,y);
        }
    }
    return 0;
}