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;
const int maxn = 200000 + 10;
const int maxf = 14;
int n, q;
int a[maxn];
LL fac[maxf + 1];
std::vector<int> get_permutation(int n, LL id) {
  bool vis[maxf + 1];
  memset(vis, false, sizeof(vis));
  std::vector<int> ans;
  for (int i = 0; i < n; i++) {
    int r = id / fac[n - 1 - i];
    id = id % fac[n - 1 - i];
    int a;
    for (a = 1; a <= maxf; a++)
      if (!vis[a]) {
        if (!r) break;
        r--;
      }
    ans.push_back(a);
    vis[a] = true;
  }
  return ans;
}
void preprocess() {
  fac[0] = 1;
  for (int i = 1; i <= maxf; i++) fac[i] = fac[i - 1] * i;
}
int main() {
  preprocess();
  scanf("%d%d", &n, &q);
  bool dirty = false;
  LL x = 0;
  if (n <= maxf) {
    std::vector<int> p;
    for (int i = 1; i <= n; i++) p.push_back(i);
    while (q--) {
      int type, l, r;
      scanf("%d%d", &type, &l);
      if (type == 1) {
        if (dirty) {
          p = get_permutation(n, x);
          dirty = false;
        }
        scanf("%d", &r);
        int ans = 0;
        for (int i = l - 1; i < r; i++) ans += p[i];
        printf("%d\n", ans);
      } else {
        x += l;
        dirty = true;
      }
    }
    return 0;
  }
  int offset = n - maxf;
  std::vector<int> p;
  for (int i = 1; i <= maxf; i++) p.push_back(i);
  while (q--) {
    int type, l, r;
    scanf("%d%d", &type, &l);
    if (type == 1) {
      if (dirty) {
        p = get_permutation(maxf, x);
        dirty = false;
      }
      scanf("%d", &r);
      LL ans = 0;
      if (l <= offset) {
        int qr = std::min(r, offset);
        ans += (LL)(l + qr) * (qr - l + 1) / 2;
      }
      if (r > offset) {
        int ql = std::max(offset + 1, l);
        for (int i = ql; i <= r; i++) ans += p[i - offset - 1];
        ans += offset * (r - ql + 1);
      }
      printf("%lld\n", ans);
    } else {
      dirty = true;
      x += l;
    }
  }
  return 0;
}