Problem Description

Recently, Doge got a funny birthday present from his new friend, Protein Tiger from St. Beeze College. No, not cactuses. It's a mysterious blackbox.

After some research, Doge found that the box is maintaining a sequence an of n numbers internally, initially all numbers are zero, and there are THREE "operations":

1.Add d to the k-th number of the sequence.

2.Query the sum of ai where l ≤ i ≤ r.

3.Change ai to the nearest Fibonacci number, where l ≤ i ≤ r.

4.Play sound "Chee-rio!", a bit useless.

Let F

₀ = 1,F

₁ = 1,Fibonacci number Fn is defined as F

_n = F

_{n - 1} + F

_{n - 2} for n ≥ 2.

Nearest Fibonacci number of number x means the smallest Fn where |F

_n - x| is also smallest.

Doge doesn't believe the machine could respond each request in less than 10ms. Help Doge figure out the reason.

Input

Input contains several test cases, please process till EOF.

For each test case, there will be one line containing two integers n, m.

Next m lines, each line indicates a query:

1 k d - "add"

2 l r - "query sum"

3 l r - "change to nearest Fibonacci"

1 ≤ n ≤ 100000, 1 ≤ m ≤ 100000, |d| < 2

³¹, all queries will be valid.

Output

For each Type 2 ("query sum") operation, output one line containing an integer represent the answer of this query.

Sample Input

1 1 2 1 1 5 4 1 1 7 1 3 17 3 2 4 2 1 5

Sample Output

0 22

Source

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解题：线段树结点含两个和，一个真正和，还有一个是3号操作的备用和。

#include
      
       #include
       
        using namespace std;struct tree{    int l,r,t;    __int64 s,ts;}ss[100010*3];__int64 f[100]={1,1};void setit(int l,int r,int d){    ss[d].l=l;    ss[d].r=r;    ss[d].t=0;//推断子结点是否须要进行3号操作    ss[d].s=0;//当前区间内的总和    ss[d].ts=1;//当前区间进行3号操作的备用和    if(l==r)        return;    setit(l,(l+r)/2,d*2);    setit((l+r)/2+1,r,d*2+1);    ss[d].ts=ss[d*2].ts+ss[d*2+1].ts;}__int64 doublekill(__int64 k)//查找与k最相近的Fibonacci{    int l=0,r=77,mid;    __int64 x1,x2;    while(l<=r)    {        mid=(l+r)/2;        if(f[mid]==k)return f[mid];        if(f[mid]
        
         0)    {        x2=f[l-1]-k;        if(x2<0)x2=-x2;        if(x2<=x1)return f[l-1];    }    return f[l];}void insert1(__int64 e,int k,int d)//点更新{    int l,r,mid;    l=ss[d].l;    r=ss[d].r;    mid=(l+r)/2;    if(l==r)    {        if(ss[d].t)        {            ss[d].s=ss[d].ts; ss[d].t=0;        }        ss[d].s+=e; ss[d].ts=doublekill(ss[d].s);//同一时候更新备用        return;    }    if(ss[d].t)//假设当前段须要进行3号操作    {        ss[d*2].t=1; ss[d*2+1].t=1; ss[d].t=0;        ss[d*2].s=ss[d*2].ts;        ss[d*2+1].s=ss[d*2+1].ts;    }    if(k>mid)insert1(e,k,d*2+1);    else insert1(e,k,d*2);    ss[d].s=ss[d*2].s+ss[d*2+1].s;    ss[d].ts=ss[d*2].ts+ss[d*2+1].ts;}void insert2(int l,int r,int d)//区间进行3号操作{    int ll,rr,mid;    ll=ss[d].l;    rr=ss[d].r;    mid=(ll+rr)/2;    if(ll>=l&&rr<=r)    {        ss[d].s=ss[d].ts; ss[d].t=1;//子结点须要3号操作更新        return;    }    if(ss[d].t)    {        ss[d*2].t=1; ss[d*2+1].t=1; ss[d].t=0;        ss[d*2].s=ss[d*2].ts;        ss[d*2+1].s=ss[d*2+1].ts;    }    if(l<=mid)insert2(l,r,d*2);    if(r>mid)insert2(l,r,d*2+1);     ss[d].s=ss[d*2].s+ss[d*2+1].s;}__int64 find(int l,int r,int d)//求和{    __int64 ll,rr,mid,s=0;    ll=ss[d].l;    rr=ss[d].r;    mid=(ll+rr)/2;    if(ll>=l&&rr<=r)    {        return ss[d].s;    }    if(ss[d].t)    {        ss[d*2].t=1; ss[d*2+1].t=1; ss[d].t=0;        ss[d*2].s=ss[d*2].ts;        ss[d*2+1].s=ss[d*2+1].ts;    }    if(l<=mid)s+=find(l,r,d*2);    if(r>mid)s+=find(l,r,d*2+1);    ss[d].s=ss[d*2].s+ss[d*2+1].s;    return s;}int main (void){    int n,m,i,j,k,l;    __int64 d;    for(i=2;i<78;i++)    f[i]=f[i-1]+f[i-2];    while(scanf("%d%d",&n,&m)>0)    {        setit(1,n,1);        while(m--)        {            scanf("%d%d",&j,&k);            if(j==1)            {                scanf("%I64d",&d);insert1(d,k,1);            }            else            {                scanf("%d",&l);                if(j==2)printf("%I64d\n",find(k,l,1));                else insert2(k,l,1);            }        }    }    return 0;}