CodeForces

Discription

By the age of three Smart Beaver mastered all arithmetic operations and got this summer homework from the amazed teacher:

You are given a sequence of integers a₁, a₂, ..., a_n. Your task is to perform on it mconsecutive operations of the following type:

1. For given numbers x_i and v_i assign value v_i to element a_xi.
2. For given numbers l_i and r_i you've got to calculate sum , where f₀ = f₁ = 1 and at i ≥ 2: f_i = f_i - 1 + f_i - 2.
3. For a group of three numbers l_i r_i d_i you should increase value a_x by d_i for allx (l_i ≤ x ≤ r_i).

Smart Beaver planned a tour around great Canadian lakes, so he asked you to help him solve the given problem.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 2·10⁵) — the number of integers in the sequence and the number of operations, correspondingly. The second line contains n integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁵). Then follow m lines, each describes an operation. Each line starts with an integer t_i (1 ≤ t_i ≤ 3) — the operation type:

• if t_i = 1, then next follow two integers x_i v_i (1 ≤ x_i ≤ n, 0 ≤ v_i ≤ 10⁵);
• if t_i = 2, then next follow two integers l_i r_i (1 ≤ l_i ≤ r_i ≤ n);
• if t_i = 3, then next follow three integers l_i r_i d_i (1 ≤ l_i ≤ r_i ≤ n, 0 ≤ d_i ≤ 10⁵).

The input limits for scoring 30 points are (subproblem E1):

• It is guaranteed that n does not exceed 100, m does not exceed 10000 and there will be no queries of the 3-rd type.

The input limits for scoring 70 points are (subproblems E1+E2):

• It is guaranteed that there will be queries of the 1-st and 2-nd type only.

The input limits for scoring 100 points are (subproblems E1+E2+E3):

• No extra limitations.

Output

For each query print the calculated sum modulo 1000000000 (10⁹).

Examples

Input

5 5
1 3 1 2 4
2 1 4
2 1 5
2 2 4
1 3 10
2 1 5

Output

12
32
8
50

Input

5 4
1 3 1 2 4
3 1 4 1
2 2 4
1 2 10
2 1 5

Output

12
45


    最近好多 数学 + 数据结构 的题啊qwq，既要动脑子又要调代码qwq
    斐波那契数列有一个非常好的性质： f[i] = f[i-1] + f[i-2] 
    这不是废话吗。。。。但是这个能推出来另一个东西 -> f[i] = f[k]*f[i-k] + f[k-1]*f[i-k-1]  (这个式子是对于f[0]=f[1]=1的斐波那契数列而言的)。
    有了这个，我们就可以随心所欲的通过 ∑f[i]*a[i+l] 和 ∑f[i+1]*a[i+l]  来构造任意  ∑f[i+d]*a[i+l] 啦(也就是程序中的Get函数)。

    然后随便打打标机(别忘了下传维护之类的)，随便合并合并子树，就可以A啦(但是要写一坨子qwq)
(最近压行成瘾qwq)

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=2*1e5,ha=1e9;
int f[maxn+5],F[maxn+5],a[maxn+5],le,ri,n,m,w,O;
int sum[maxn*4+5][2],tag[maxn*4+5],L[maxn*4+5],A;
inline void ADD(int &x,int y){ x+=y; if(x>=ha) x-=ha;}
inline int add(int x,int y){ x+=y; return x>=ha?x-ha:x;}
inline void init(){ f[0]=F[0]=1; for(int i=1;i<=maxn;i++) f[i]=add(f[i-1],f[i-2]),F[i]=add(f[i],F[i-1]);}
inline int Get(int o,int Derta){ return Derta<2?(Derta?sum[o][1]:sum[o][0]):add(sum[o][0]*(ll)f[Derta-2]%ha,sum[o][1]*(ll)f[Derta-1]%ha);}
inline void QADD(int o,int Derta){ ADD(tag[o],Derta),ADD(sum[o][0],Derta*(ll)F[L[o]-1]%ha),ADD(sum[o][1],Derta*(ll)add(F[L[o]],ha-1)%ha);}
inline void pushdown(int o,int lc,int rc){ if(tag[o]){ QADD(lc,tag[o]),QADD(rc,tag[o]),tag[o]=0;}}
inline void Merge(int o,int lc,int rc,int Derta){
	sum[o][0]=add(sum[lc][0],Get(rc,Derta));
	sum[o][1]=add(sum[lc][1],Get(rc,Derta+1));
}

void build(int o,int l,int r){
	L[o]=r-l+1;
	if(l==r){ sum[o][0]=sum[o][1]=a[l]; return;}
	int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;
	build(lc,l,mid);
	build(rc,mid+1,r);
	Merge(o,lc,rc,mid+1-l);
}

void update1(int o,int l,int r){
	if(l==r){ sum[o][0]=sum[o][1]=w; return;}
	int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;
	pushdown(o,lc,rc);
	if(le<=mid) update1(lc,l,mid);
	else update1(rc,mid+1,r);
	Merge(o,lc,rc,mid+1-l);
}

void update2(int o,int l,int r){
	if(l>=le&&r<=ri){ QADD(o,w); return;}
	int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;
	pushdown(o,lc,rc);
	if(le<=mid) update2(lc,l,mid);
	if(ri>mid) update2(rc,mid+1,r);
	Merge(o,lc,rc,mid+1-l);	
}

void query(int o,int l,int r){
	if(l>=le&&r<=ri){ ADD(A,Get(o,l-le)); return;}
	int mid=l+r>>1,lc=o<<1,rc=(o<<1)|1;
	pushdown(o,lc,rc);
	if(le<=mid) query(lc,l,mid);
	if(ri>mid) query(rc,mid+1,r);
}

inline void solve(){
	while(m--){
		scanf("%d",&O);
		if(O==1){ scanf("%d%d",&le,&w),update1(1,1,n);}
		else if(O==2){ scanf("%d%d",&le,&ri),A=0,query(1,1,n),printf("%d
",A);}
		else{ scanf("%d%d%d",&le,&ri,&w),update2(1,1,n);}
	}
}

int main(){
	init();
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) scanf("%d",a+i);
	build(1,1,n);
	solve();
	return 0;
}