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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N, M, a[200001];
long long c[200001], sum, fc[20];
bool vis[20];
int get() {
  int x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x;
}
void put(long long x) {
  if (x >= 10) put(x / 10);
  putchar(x % 10 + '0');
}
void upd(int x, int v) {
  for (int i = x; i <= N; i += i & -i) c[i] += v;
}
long long ask(int x) {
  long long ans = 0;
  for (int i = x; i; i -= i & -i) ans += c[i];
  return ans;
}
void upd1(long long x) {
  for (int i = 1; i <= N; i++) vis[i] = 0;
  for (int i = 1; i <= N; i++) {
    int id = 0, cnt = 0;
    while (cnt <= x / fc[N - i]) {
      id++;
      while (vis[id]) id++;
      cnt++;
    }
    vis[id] = 1, upd(i, id - a[i]), a[i] = id, x %= fc[N - i];
  }
}
void upd2(long long x) {
  for (int i = 1; i <= 15; i++) vis[i] = 0;
  for (int i = 1; i <= 15; i++) {
    int id = 0, cnt = 0;
    while (cnt <= x / fc[15 - i]) {
      id++;
      while (vis[id]) id++;
      cnt++;
    }
    vis[id] = 1, upd(N - 15 + i, N - 15 + id - a[N - 15 + i]),
    a[N - 15 + i] = N - 15 + id, x %= fc[15 - i];
  }
}
void sol1() {
  for (int i = 1; i <= M; i++) {
    int opt = get(), x = get(), y;
    opt == 1 ? put(ask(y = get()) - ask(x - 1)), putchar('\n'),
        0    : (upd1(sum += x), 0);
  }
}
void sol2() {
  for (int i = 1; i <= M; i++) {
    int opt = get(), x = get(), y;
    opt == 1 ? put(ask(y = get()) - ask(x - 1)), putchar('\n'),
        0    : (upd2(sum += x), 0);
  }
}
int main() {
  N = get(), M = get();
  for (int i = 1; i <= N; i++) a[i] = i, upd(i, i);
  fc[0] = 1;
  for (int i = 1; i <= 15; i++) fc[i] = fc[i - 1] * i;
  N <= 15 ? sol1() : sol2();
  return 0;
}