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;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const long long linf = 4000000000000000000LL;
const long long inf = 1000000007;
const long double pi = 3.1415926535;
void pv(vector<int> a) {
  for (auto& x : a) cout << x << " ";
  cout << '\n';
}
void pv(vector<long long> a) {
  for (auto& x : a) cout << x << " ";
  cout << '\n';
}
void pv(vector<vector<int> > a) {
  for (int i = (0); i < (int(a.size())); ++i) {
    cout << i << '\n';
    pv(a[i]);
    cout << '\n';
  }
}
void pv(vector<vector<long long> > a) {
  for (int i = (0); i < (int(a.size())); ++i) {
    cout << i << '\n';
    pv(a[i]);
  }
  cout << '\n';
}
void pv(vector<string> a) {
  for (auto& x : a) cout << x << '\n';
  cout << '\n';
}
void setIO(string second) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (int(second.size())) {
    freopen((second + ".in").c_str(), "r", stdin);
    if (second != "test2") freopen((second + ".out").c_str(), "w", stdout);
  }
}
int n, q;
vector<int> a;
int SZ = 14;
int start;
long long cur;
vector<long long> fac;
void init() {
  vector<int> index(SZ);
  long long t = cur;
  for (int i = (0); i < (SZ); ++i) {
    int size = SZ - i - 1;
    index[i] = t / fac[size] + 1;
    t %= fac[size];
  }
  vector<int> visited(SZ);
  for (int i = (0); i < (SZ); ++i) {
    int p = 0;
    while (index[i]) {
      if (!visited[p]) index[i]--;
      if (index[i]) p++;
    }
    visited[p] = 1;
    a[i] = p;
  }
  for (auto& x : a) x += start;
}
long long solve(int i) {
  long long ans = 0;
  if (i >= start - 1) {
    ans = ((long long)(start - 1) * (start)) / 2;
    int j = i - start + 2;
    for (int k = (0); k < (j); ++k) ans += a[k];
  } else {
    ans = ((long long)(i + 1)) * (i + 2) / 2;
  }
  return ans;
}
int main() {
  setIO("");
  cin >> n >> q;
  SZ = min(SZ, n);
  fac.resize(SZ + 1);
  fac[0] = 1;
  for (int i = (0); i < (SZ); ++i) fac[i + 1] = fac[i] * (i + 1);
  a.resize(SZ);
  start = n - SZ + 1;
  cur = 0;
  init();
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r;
      cin >> l >> r;
      cout << solve(r - 1) - solve(l - 2) << '\n';
    } else {
      int x;
      cin >> x;
      cur += x;
      init();
    }
  }
}