Problem A
Adventure Begins

The game Pokenom Go has just been released. Pokenom trainers can now travel the world, capture Pokenom in the wild and battle each other! Bash — the Pokenom trainer — has decided to drop out of his university to pursue his childhood dream of becoming the best Pokenom trainer!

However, Linux — Bash’s university headmaster — does not allow his students to drop out so easily …

Linux puts $N$ black boxes on a straight line. The black boxes are numbered from $1$ to $N$ from left to right. Initially, all black boxes are empty. Then Linux gives Bash $Q$ queries. Each query can be one of the following $2$ types:

• Linux puts exactly one stone inside exactly one box between $u$-th box and $v$-th box, inclusive, with equal probability. $(1 \le u \le v \le N)$.

• Let $a_ i$ be the number of stones in black box numbered $i$. Let $A = \sum _{i=1}^{N}{a_ i^2}$. Bash has to calculate the expected value $E(A)$.

Bash can only drop out of his university if he is able to answer all queries correctly. But now all Bash can think of is Pokenom. Please help him!

Input

The first line of input contains exactly $2$ positive integers $N$ and $Q$. $(1 \le N, Q \le 10^5)$.

$Q$ lines follow, each line contains exactly one query. As explained, a query can be one of the following $2$ types:

• $1 \; u \; v$: Linux puts a stone inside one of the boxes between $u$ and $v$.

• $2$: Linux asks Bash to compute $E(A)$.

Output

It can be proved that the expected value can be represented as an irreducible fraction $\dfrac {A}{B}$. For each query of type $2$, print one line containing the value $A \times B^{-1}$ modulo $10^{9} + 7$. The given input guarantees that $B$ is not a multiple of $10^{9} + 7$.

Explanation for examples

• In the first example: With a probability of $0.5$, two stones are in different squares. Hence, the answer to the fourth query is $0.5 \times (1^{2} + 1^{2}) + 0.5 \times 2^{2} = 3$.

• In the second example: With a probability of $\frac{2}{3}$, two stones are in different squares. Hence, the answer to the fourth query is $\frac{2}{3} \times 2 + \frac{1}{3} \times 4 = \frac{8}{3}$.

2 4
1 1 2
2
1 1 2
2

1
3

3 4
1 1 3
2
1 1 3
2

1
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