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;
vector<long long> fact(16);
void solve() {
  long long n, q, cnt(0);
  cin >> n >> q;
  long long len = min(n, 15ll);
  vector<long long> v(len), a(n);
  vector<bool> vis;
  iota(a.begin(), a.end(), n - len + 1);
  iota(v.begin(), v.end(), n - len + 1);
  function<long long(long long)> kmax = [&](long long k) {
    priority_queue<pair<long long, long long>> q;
    long long sz(0);
    for (long long i = 0; i < len; i++) {
      if (vis[i]) continue;
      if (sz < k)
        q.push({a[i], i}), sz++;
      else
        q.push({a[i], i}), q.pop();
    }
    return q.top().second;
  };
  function<void(long long)> update = [&](long long cnt) {
    vis.assign(len, 0);
    for (long long i = len - 1; i >= 0; i--) {
      long long q = cnt / fact[i];
      cnt = cnt % fact[i];
      long long idx = kmax(q + 1);
      v[len - i - 1] = a[idx];
      vis[idx] = 1;
    }
  };
  while (q--) {
    long long t;
    cin >> t;
    if (t - 1) {
      long long x;
      cin >> x;
      cnt += x;
      update(cnt);
    } else {
      long long l, r;
      cin >> l >> r;
      long long s1(0), s2(0);
      if (n <= 15) {
        l--, r--;
        for (long long i = l; i <= r; i++) s1 += v[i];
      } else {
        long long s = max(0ll, len - n + l - 1);
        long long e = r + len - n;
        r = min(a[0] - 1, r);
        l--;
        if (l < r)
          s1 = ((((long long)r * (long long)(r + 1)) >> 1ll) -
                (((long long)l * (long long)(l + 1)) >> 1ll));
        for (; s < e; s++) s2 += v[s];
      }
      cout << s1 + s2 << "\n";
    }
  }
}
signed main() {
  fact[0] = 1;
  for (long long i = 1; i < 16; i++) fact[i] = fact[i - 1] * i;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t(1);
  for (long long i = 1; i <= t; solve(), i++)
    ;
  return 0;
}