Problem: A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] — are permutations, and [1, 1], [4, 3, 1], [2, 3, 4] — no.

Permutation a is lexicographically smaller than permutation b (they have the same length n), if in the first index i in which they differ, a[i] < b[i]. For example, the permutation [1, 3, 2, 4] is lexicographically smaller than the permutation [1, 3, 4, 2], because the first two elements are equal, and the third element in the first permutation is smaller than in the second.

The next permutation for a permutation a of length n — is the lexicographically smallest permutation b of length n that lexicographically larger than a. For example: 

  * for permutation [2, 1, 4, 3] the next permutation is [2, 3, 1, 4]; 
  * for permutation [1, 2, 3] the next permutation is [1, 3, 2]; 
  * for permutation [2, 1] next permutation does not exist. 



You are given the number n — the length of the initial permutation. The initial permutation has the form a = [1, 2, …, n]. In other words, a[i] = i (1 ≤ i ≤ n).

You need to process q queries of two types: 

  * 1 l r: query for the sum of all elements on the segment [l, r]. More formally, you need to find a[l] + a[l + 1] + … + a[r]. 
  * 2 x: x times replace the current permutation with the next permutation. For example, if x=2 and the current permutation has the form [1, 3, 4, 2], then we should perform such a chain of replacements [1, 3, 4, 2] → [1, 4, 2, 3] → [1, 4, 3, 2]. 



For each query of the 1-st type output the required sum.

Input

The first line contains two integers n (2 ≤ n ≤ 2 ⋅ 10^5) and q (1 ≤ q ≤ 2 ⋅ 10^5), where n — the length of the initial permutation, and q — the number of queries.

The next q lines contain a single query of the 1-st or 2-nd type. The 1-st type query consists of three integers 1, l and r (1 ≤ l ≤ r ≤ n), the 2-nd type query consists of two integers 2 and x (1 ≤ x ≤ 10^5).

It is guaranteed that all requests of the 2-nd type are possible to process.

Output

For each query of the 1-st type, output on a separate line one integer — the required sum.

Example

Input


4 4
1 2 4
2 3
1 1 2
1 3 4


Output


9
4
6

Note

Initially, the permutation has the form [1, 2, 3, 4]. Queries processing is as follows: 

  1. 2 + 3 + 4 = 9; 
  2. [1, 2, 3, 4] → [1, 2, 4, 3] → [1, 3, 2, 4] → [1, 3, 4, 2]; 
  3. 1 + 3 = 4; 
  4. 4 + 2 = 6 

Solution (in PYTHON3):

Solution:
from math import factorial

def short_perm_to_num(perm, n): # O(n**2), but n <= 70
    result = 0
    is_used = {}
    for i in range(n):
        for j in range(perm[i]):
            if not (j in is_used):
                result += factorial(n-1-i)
        is_used[perm[i]] = True
    return result

def perm_to_num(perm, n):
    max_len = 15
    if n <= max_len:
        return short_perm_to_num(perm, n)
    else:
        last_part = [x - (n-max_len) for x in perm[n-max_len:]]
        return short_perm_to_num(last_part, max_len)

def short_num_to_perm(num, n):
    is_used = {}
    perm = []
    for i in range(n):
        deg = num // factorial(n-1-i)
        num %= factorial(n-1-i)
        j = 0
        while j in is_used:
            j += 1
        # now j is 1st unused item
        deg_of_j = 0
        while (deg_of_j < deg) or (j in is_used):
            if not (j in is_used):
                deg_of_j += 1
            j += 1
        perm.append(j)
        is_used[j] = True
    return perm

def num_to_perm(num, n): # RETURNS ONLY LAST 15 SYMBOLS
    max_len = 15
    if n <= max_len:
        return short_num_to_perm(num, n)
    else:
        last_part_low = short_num_to_perm(num, max_len)
        last_part = [x + (n-max_len) for x in last_part_low]
        return last_part


[n, q] = [int(x) for x in input().split()]
full_perm = [i for i in range(0, n)] # full permutation as an array, but don't forget to add 1 when printing
num_of_perm = 0

# 70! is just above 10**100, so elements before the last 70 are still 1, 2, 3, ... n-7


for quarry_num in range(q):
    quarry = [int(x) for x in input().split()]
    if quarry[0] == 1:
        perm = num_to_perm(num_of_perm, n)
        l = quarry[1]
        r = quarry[2]
        if n <= 15:
            print(sum(perm[l-1:r]) + r-l+1)
        else: # n-15, n-14, ... n-1 are in the perm; we must compute sum(perm[l-1:r])
            if l-1 >= n-15:
                print(sum(perm[l-1-n+15:r-n+15]) + r-l+1)
            else: # l-1 < n-15
                if r < n-15:
                    s = ((l-1) + r-1) * (r-1 - l+1 + 1) // 2
                    print(s + r-l+1)
                else:
                    s = ((l-1) + n-16) * (n-16 - l + 2) // 2
                    rest = sum(perm[:r-n+15])
                    print(s + rest + r-l+1)
    if quarry[0] == 2:
        num_of_perm += quarry[1]