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;
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p);
template <class T>
ostream &operator<<(ostream &os, vector<T> &v);
template <class T>
ostream &operator<<(ostream &os, set<T> &v);
template <class T1, class T2>
ostream &operator<<(ostream &os, map<T1, T2> &v);
const int N = 3e5 + 5;
const long long mod = 998244353;
vector<long long> prefix;
long long get_sum(long long l, long long r) {
  if (l > r) return 0;
  long long ret = prefix[r];
  if (l > 0) ret -= prefix[l - 1];
  return ret;
}
long long get_sum2(long long l, long long r, vector<long long> &v) {
  long long ret = 0;
  for (long long i = l; i <= r; i++) {
    ret += v[i];
  }
  return ret;
}
vector<long long> sz;
void fun(vector<long long> &p, long long now, vector<long long> &ret,
         long long len) {
  if (ret.size() == len) {
    return;
  }
  vector<long long> v;
  {
    for (long long i = 0; i < p.size(); i++) {
      if (find(ret.begin(), ret.end(), p[i]) == ret.end()) {
        v.push_back(p[i]);
      }
    }
    sort(v.begin(), v.end());
  }
  long long lagbe = len - ret.size();
  long long ache = sz[lagbe - 1];
  assert(v.size() == lagbe);
  for (long long i = 0; i < lagbe; i++) {
    if (ache > now) {
      ret.push_back(v[i]);
      fun(p, now, ret, len);
      break;
    } else {
      now -= ache;
    }
  }
}
void TEST_CASES(int cas) {
  vector<long long> v;
  long long n, q;
  cin >> n >> q;
  prefix.resize(n);
  for (long long i = 1; i <= n; i++) {
    v.push_back(i);
  }
  for (long long i = 0; i < n; i++) {
    prefix[i] = (i - 1 >= 0 ? prefix[i - 1] : 0) + v[i];
  };
  long long koto = min(14ll, n);
  long long now = 0;
  for (long long i = 0; i <= koto; i++) {
    if (i == 0) {
      sz.push_back(1);
      continue;
    }
    sz.push_back(i * sz.back());
  };
  while (q--) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long l, r;
      cin >> l >> r;
      l--;
      r--;
      vector<long long> p;
      for (long long i = 0; i < koto; i++) {
        p.push_back(n - koto + 1 + i);
      }
      vector<long long> ans;
      fun(p, now, ans, koto);
      ;
      long long sum1 = 0, sum2 = 0;
      {
        long long x = n - koto;
        long long y = koto;
        if (r >= x) {
          sum1 += get_sum(0, x - 1);
          ;
          r -= x;
          sum1 += get_sum2(0, r, ans);
          ;
        } else {
          sum1 += get_sum(0, r);
        }
      }
      l--;
      {
        long long x = n - koto;
        long long y = koto;
        if (l >= x) {
          sum2 += get_sum(0, x - 1);
          l -= x;
          sum2 += get_sum2(0, l, ans);
        } else {
          sum2 += get_sum(0, l);
        }
      };
      cout << sum1 - sum2 << "\n";
    } else {
      long long x;
      cin >> x;
      now += x;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1, cas = 0;
  while (t--) {
    TEST_CASES(++cas);
  }
  return 0;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
  os << "{" << p.first << ", " << p.second << "} ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> &v) {
  os << "[ ";
  for (int i = 0; i < v.size(); i++) {
    os << v[i] << " ";
  }
  os << " ]";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, set<T> &v) {
  os << "[ ";
  for (T i : v) {
    os << i << " ";
  }
  os << " ]";
  return os;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, map<T1, T2> &v) {
  for (auto i : v) {
    os << "Key : " << i.first << " , Value : " << i.second << endl;
  }
  return os;
}