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;
long long n, q;
const long long MXN = 200005;
long long arr[MXN];
long long seg[MXN * 4];
void build(long long ind, long long l, long long r) {
  if (l == r) {
    seg[ind] = arr[l];
    return;
  }
  long long m = (l + r) / 2;
  build(ind << 1, l, m);
  build((ind << 1) | 1, m + 1, r);
  seg[ind] = seg[ind * 2] + seg[ind * 2 + 1];
}
void upd(long long ind, long long l, long long r, long long pos, long long to) {
  if (l == r) {
    seg[ind] = to;
    return;
  }
  long long m = (l + r) / 2;
  if (pos <= m)
    upd(ind << 1, l, m, pos, to);
  else
    upd((ind << 1) | 1, m + 1, r, pos, to);
  seg[ind] = seg[ind * 2] + seg[ind * 2 + 1];
}
long long qry(long long ind, long long l, long long r, long long ql,
              long long qr) {
  if (ql <= l && r <= qr) {
    return seg[ind];
  }
  if (ql > r || qr < l) {
    return 0;
  }
  long long m = (l + r) / 2;
  return qry(ind << 1, l, m, ql, qr) + qry((ind << 1) | 1, m + 1, r, ql, qr);
}
long long fact[MXN];
set<long long> rm;
long long T(long long dv) {
  auto x = rm.begin();
  for (long long j = 0; j < dv; j++) x++;
  return *x;
}
vector<long long> wh;
void sep(long long kth) {
  for (long long i = 1; i <= n; i++) {
    rm.insert(i);
  }
  for (long long pl = n - 1; pl >= 0; pl--) {
    long long dv = kth / fact[pl];
    kth %= fact[pl];
    long long K = T(dv);
    wh.push_back(K);
    rm.erase(K);
  }
  for (long long i = 1; i <= n; i++) {
    arr[i] = wh[i - 1];
  }
  wh.clear();
}
long long SP = 0;
void rew(long long kth) {
  if (SP == kth) return;
  if (n <= 15) {
    sep(kth);
  } else {
    for (long long i = 1; i <= 15; i++) {
      rm.insert(i);
    }
    for (long long pl = 15 - 1; pl >= 0; pl--) {
      long long dv = kth / fact[pl];
      kth %= fact[pl];
      long long K = T(dv);
      wh.push_back(K);
      rm.erase(K);
    }
    for (long long i = 1; i <= 15; i++) {
      arr[n - 15 + i] = n - 15 + wh[i - 1];
    }
    wh.clear();
  }
  SP = kth;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  fact[0] = 1;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) arr[i] = i;
  for (long long i = 1; i <= 20; i++) {
    fact[i] = fact[i - 1] * i;
  }
  long long k = 0;
  build(1, 1, n);
  for (long long i = 0; i < q; i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      rew(k);
      for (long long i = max(1ll, n - 20); i <= n; i++) {
        upd(1, 1, n, i, arr[i]);
      }
      long long l, r;
      cin >> l >> r;
      cout << qry(1, 1, n, l, r) << '\n';
    } else {
      long long w;
      cin >> w;
      k += w;
    }
  }
}