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;
struct segtree {
  int sz;
  vector<long long> sums;
  void init(int n) {
    sz = n;
    sums.resize(n << 2);
  }
  void build(vector<long long> &v, int nd, int l, int r) {
    if (l == r) {
      sums[nd] = v[l];
      return;
    }
    int mid = (l + r) / 2;
    build(v, nd << 1, l, mid);
    build(v, nd << 1 | 1, mid + 1, r);
    sums[nd] = sums[nd << 1] + sums[nd << 1 | 1];
  }
  void build(vector<long long> &v) { build(v, 1, 1, sz); }
  void update(int nd, int l, int r, int pos, int val) {
    if (l == pos && r == pos) {
      sums[nd] = val;
      return;
    }
    int mid = (l + r) / 2;
    if (pos <= mid)
      update(nd << 1, l, mid, pos, val);
    else
      update(nd << 1 | 1, mid + 1, r, pos, val);
    sums[nd] = sums[nd << 1] + sums[nd << 1 | 1];
  }
  void update(int pos, int val) { update(1, 1, sz, pos, val); }
  long long querysum(int nd, int l, int r, int from, int to) {
    if (from <= l && r <= to) return sums[nd];
    if (from > r || l > to) return 0LL;
    int mid = (l + r) / 2;
    return querysum(nd << 1, l, mid, from, to) +
           querysum(nd << 1 | 1, mid + 1, r, from, to);
  }
  long long querysum(int l, int r) { return querysum(1, 1, sz, l, r); }
};
const int N = 200100;
long long n, sum, m, fact[21];
vector<long long> a;
segtree st;
set<long long> s;
int getNum(set<long long> &s, int num) {
  for (auto &x : s) {
    if (num == 0) return x;
    num--;
  }
  assert(0);
}
vector<long long> get(long long sum) {
  int cnt = 0;
  s.clear();
  for (long long i = n; i >= max(n - 20 + 1, 1LL); i--) {
    s.insert(i);
    cnt++;
  }
  vector<long long> v;
  while (cnt--) {
    int num = sum / fact[cnt];
    sum -= num * fact[cnt];
    int setter = getNum(s, num);
    v.push_back(setter);
    if (s.find(setter) != s.end()) s.erase(setter);
  }
  return v;
}
int main() {
  fact[0] = 1;
  for (int i = 1; i < 21; i++) fact[i] = fact[i - 1] * i;
  scanf("%lld%lld", &n, &m);
  a.resize(n + 10);
  for (int i = 1; i <= n; i++) a[i] = i;
  st.init(n);
  st.build(a);
  while (m--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%lld\n", st.querysum(l, r));
    } else {
      long long x;
      scanf("%lld", &x);
      sum += x;
      vector<long long> v = get(sum);
      int sz = (int)(v.size());
      int cnt = 0;
      for (int i = n - sz + 1; i <= n; i++) {
        st.update(i, v[cnt]);
        cnt++;
      }
    }
  }
}