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 ll = long long;
using ull = unsigned long long;
constexpr int INF = 1073741823;
constexpr int MINF = -1073741823;
constexpr ll LINF = ll(4661686018427387903);
constexpr ll MOD = 1e9 + 7;
constexpr ll mod = 998244353;
constexpr long double eps = 1e-6;
const long double PI = acosl(-1.0L);
using namespace std;
void scans(string& str) {
  char c;
  str = "";
  scanf("%c", &c);
  if (c == '\n') scanf("%c", &c);
  while (c != '\n' && c != -1 && c != ' ') {
    str += c;
    scanf("%c", &c);
  }
}
void scanc(char& str) {
  char c;
  scanf("%c", &c);
  if (c == -1) return;
  while (c == '\n') {
    scanf("%c", &c);
  }
  str = c;
}
double acot(double x) { return PI / 2 - atan(x); }
ll LSB(ll n) { return (n & (-n)); }
template <typename T>
inline T chmin(T& a, const T& b) {
  if (a > b) a = b;
  return a;
}
template <typename T>
inline T chmax(T& a, const T& b) {
  if (a < b) a = b;
  return a;
}
template <typename T = int>
vector<T> kth_permutation(int n, T k) {
  vector<T> a(n), b(n);
  for (int i = 0; i < (n); ++i) b[i] = i + 1;
  vector<T> x(n);
  for (int i = 0; i < (n); ++i) {
    x[i] = (k % (i + 1));
    k /= (i + 1);
  }
  reverse((x).begin(), (x).end());
  for (int i = 0; i < (n); ++i) {
    a[i] = b[x[i]];
    b.erase(b.begin() + x[i]);
  }
  return a;
}
int main() {
  ll n, qq = 0;
  int q;
  scanf("%lld%d", &n, &q);
  if (n < 15) {
    vector<int> a(n);
    for (int i = 0; i < (n); ++i) a[i] = i + 1;
    while (q--) {
      int t;
      scanf("%d", &t);
      if (t == 1) {
        int l, r;
        scanf("%d%d", &l, &r);
        int ans = 0;
        for (int i = l - 1; i < r; ++i) ans += a[i];
        printf("%d\n", ans);
      } else {
        int p;
        scanf("%d", &p);
        while (p--) next_permutation((a).begin(), (a).end());
      }
    }
  } else {
    vector<ll> a(15);
    for (int i = 0; i < (15); ++i) a[i] = n - i;
    reverse((a).begin(), (a).end());
    while (q--) {
      int t;
      scanf("%d", &t);
      if (t == 1) {
        ll l, r;
        scanf("%lld%lld", &l, &r);
        ll ans = 0;
        if (l > n - 15) {
          l -= n - 15 + 1;
          r -= n - 15;
          while (l < r) {
            ans += a[l];
            ++l;
          }
        } else if (r > n - 15) {
          ans += (l + n - 15) * (n - 15 - l + 1) / 2;
          for (int i = 0; i < r - (n - 15); ++i) ans += a[i];
        } else {
          ans += (l + r) * (r - l + 1) / 2;
        }
        printf("%lld\n", ans);
      } else {
        ll p;
        scanf("%lld", &p);
        qq += p;
        p = qq;
        a = kth_permutation(15, p);
        for (int i = 0; i < (15); ++i) a[i] += n - 15;
      }
    }
  }
  return 0;
}