C. Sasha and Array time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output

Sasha has an array of integers a₁, a₂, ..., a_n. You have to perform m queries. There might be queries of two types:

1. 1 l r x — increase all integers on the segment from l to r by values x;
2. 2 l r — find , where f(x) is the x-th Fibonacci number. As this number may be large, you only have to find it modulo 10⁹
    + 7.

In this problem we define Fibonacci numbers as follows: f(1) = 1, f(2) = 1, f(x) = f(x - 1) + f(x - 2) for all x > 2.

Sasha is a very talented boy and he managed to perform all queries in five seconds. Will you be able to write the program that performs as well as Sasha?

Input

The first line of the input contains two integers n

and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — the number of elements in the array and the number of queries respectively.

The next line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Then follow m lines with queries descriptions. Each of them contains integers tp_i, l_i, r_i and may be x_i (1 ≤ tp_i ≤ 2, 1 ≤ l_i

 ≤ r_i ≤ n, 1 ≤ x_i ≤ 10⁹). Here tp_i = 1 corresponds to the queries of the first type and tp_i corresponds to the queries of the second type.

It's guaranteed that the input will contains at least one query of the second type.

Output

For each query of the second type print the answer modulo 10⁹ + 7.

Examples Input

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

Output

5
7
9

Note

Initially, array a is equal to 1, 1, 2, 1, 1.

The answer for the first query of the second type is f(1) + f(1) + f(2) + f(1) + f(1) = 1 + 1 + 1 + 1 + 1 = 5.

After the query 1 2 4 2 array a is equal to 1, 3, 4, 3, 1.

The answer for the second query of the second type is f(3) + f(4) + f(3) = 2 + 3 + 2 = 7.

The answer for the third query of the second type is f(1) + f(3) + f(4) + f(3) + f(1) = 1 + 2 + 3 + 2 + 1 = 9.

題目大意：

    有一棵線段樹，查詢以每個端點為下標值的斐波那契值的和。

解題思路：

    一般情況下，求任意斐波那契數我們使用的是矩陣快速冪，這題是一個對於斐波那契數的線段樹，所以我們線上段樹上維護矩陣即可。

AC程式碼（用了Q巨的模版）：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;
#define MOD 1000000007

const int MAXN=100000+3;

int a[MAXN];
struct Matrix
{
    int a[2][2];//矩陣大小根據需求修改
    Matrix()
    {
        memset(a,0,sizeof(a));
    }
    void init()
    {
        for(int i=0;i<2;i++)
            for(int j=0;j<2;j++)
                a[i][j]=(i==j);
    }
    Matrix operator + (const Matrix &B)const
    {
        Matrix C;
        for(int i=0;i<2;i++)
            for(int j=0;j<2;j++)
                C.a[i][j]=(a[i][j]+B.a[i][j])%MOD;
        return C;
    }
    Matrix operator * (const Matrix &B)const
    {
        Matrix C;
        for(int i=0;i<2;i++)
            for(int k=0;k<2;k++)
                for(int j=0;j<2;j++)
                    C.a[i][j]=(C.a[i][j]+1LL*a[i][k]*B.a[k][j])%MOD;
        return C;
    }
    Matrix operator ^ (const int &t)const
    {
        Matrix A=(*this),res;
        res.init();
        int p=t;
        while(p)
        {
            if(p&1)res=res*A;
            A=A*A;
            p>>=1;
        }
        return res;
    }
};
Matrix f,s[MAXN<<2],lz[MAXN<<2];
void pushUp(int n)
{
    s[n]=s[n<<1]+s[n<<1|1];
}
void pushDown(int n)
{
    s[n<<1]=lz[n]*s[n<<1];
    s[n<<1|1]=lz[n]*s[n<<1|1];
    lz[n<<1]=lz[n]*lz[n<<1];
    lz[n<<1|1]=lz[n]*lz[n<<1|1];
    lz[n].init();
}
void build(int l,int r,int n)//使用時輸入(1,N,1)，N為總長度
{
    lz[n].init();
    if(l==r)
    {
        s[n]=f^(a[l]-1);
        return;
    }
    int m=(l+r)/2;
    build(l,m,n<<1);
    build(m+1,r,n<<1|1);
    pushUp(n);
}
void update(int ql,int qr,int l,int r,Matrix v,int n)//使用時輸入(l,r.1,N,x,1)
{
    if(l==ql && r==qr)
    {
        s[n]=v*s[n];
        lz[n]=v*lz[n];
        return;
    }
    pushDown(n);
    int m=(l+r)/2;
    if(qr<=m)
        update(ql,qr,l,m,v,n<<1);
    else if(ql>m)
        update(ql,qr,m+1,r,v,n<<1|1);
    else
    {
        update(ql,m,l,m,v,n<<1);
        update(m+1,qr,m+1,r,v,n<<1|1);
    }
    pushUp(n);
}
int query(int ql,int qr,int l,int r,int n)
{
    if(l==ql && r==qr)
        return s[n].a[0][0];
    pushDown(n);
    int m=(l+r)/2;
    if(qr<=m)
        return query(ql,qr,l,m,n<<1);
    if(ql>m)
        return query(ql,qr,m+1,r,n<<1|1);
    return (query(ql,m,l,m,n<<1)+query(m+1,qr,m+1,r,n<<1|1))%MOD;
}

int N,M;

int main()
{
    f.a[0][0]=1; f.a[0][1]=1;
    f.a[1][0]=1; f.a[1][1]=0;
    scanf("%d%d",&N,&M);
    for(int i=1;i<=N;++i)
        scanf("%d",&a[i]);
    build(1,N,1);
    while(M--)
    {
        int type,l,r;
        scanf("%d%d%d",&type,&l,&r);
        if(type==1)
        {
            int x;
            scanf("%d",&x);
            update(l,r,1,N,f^x,1);
        }
        else printf("%d\n",query(l,r,1,N,1));
    }
    
    return 0;
}