CF1599E.Two Arrays

You are given two integer arrays of length $N$ , $A1$ and $A2$ . You are also given $Q$ queries of 4 types:

1 k l r x: set $Ak_i:=min(Ak_i, x)$ for each $l \leq i \leq r$ .

2 k l r x: set $Ak_i:=max(Ak_i, x)$ for each $l \leq i \leq r$ .

3 k l r x: set $Ak_i:=Ak_i+x$ for each $l \leq i \leq r$ .

4 l r: find the $(\sum_{i=l}^r F(A1_i+A2_i)) \% (10^9+7)$ where $F(k)$ is the $k$ -th Fibonacci number ( $F(0)=0, F(1)=1, F(k)=F(k-1)+F(k-2)$ ), and $x \% y$ denotes the remainder of the division of $x$ by $y$ .

You should process these queries and answer each query of the fourth type.

The first line contains two integers $N$ and $Q$ . ( $1 \leq N, Q \leq 5 \times 10^4$ )

The second line contains $N$ integers, array $A1_1, A1_2, \dots A1_N$ . ( $0 \leq A1_i \leq 10^6$ )

The third line contains $N$ integers, array $A2_1, A2_2, \dots A2_N$ . ( $0 \leq A2_i \leq 10^6$ )

The next $Q$ lines describe the queries. Each line contains 5 or 3 integers, where the first integer denotes the type of the query. ( $k \in \{1, 2\}$ , $1 \leq l \leq r \leq N$ )

For queries of type 1 and 2, $0 \leq x \leq 10^9$ holds.

For queries of type 3, $−10^6 \leq x \leq 10^6$ holds.

It is guaranteed that after every query each number in arrays $A1$ and $A2$ will be nonnegative.

Print the answer to each query of the fourth type, in separate lines.

In the first example: The answer for the first query is $F(1 + 2) + F(0 + 1) + F(2 + 0) = F(3) + F(1) + F(2) = 2 + 1 + 1 = 4$ . After the second query, the array $A2$ changes to $[2, 4, 0]$ . After the third query, the array $A1$ changes to $[0, 0, 0]$ . The answer for the fourth query is $F(0 + 2) + F(0 + 4) + F(0 + 0) = F(2) + F(4) + F(0) = 1 + 3 + 0 = 4$ .

In the second example: The answer for the first query is $F(1 + 4) + F(3 + 2) + F(5 + 1) = F(5) + F(5) + F(6) = 5 + 5 + 8 = 18$ . The answer for the second query is $F(3 + 2) + F(5 + 1) + F(3 + 3) + F(2 + 3) = F(5) + F(6) + F(6) + F(5) = 5 + 8 + 8 + 5 = 26$ . After the third query, the array $A1$ changes to $[1, 6, 6, 6, 2]$ . The answer for the fourth query is $F(6 + 2) + F(6 + 1) + F(6 + 3) = F(8) + F(7) + F(9) = 21 + 13 + 34 = 68$ .