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;
inline int add(int _a, int _b) {
  if (_a < 0) {
    _a += 1000000007;
  }
  if (_b < 0) {
    _b += 1000000007;
  }
  if (_a + _b >= 1000000007) {
    return _a + _b - 1000000007;
  }
  return _a + _b;
}
inline int mul(int _a, int _b) {
  if (_a < 0) {
    _a += 1000000007;
  }
  if (_b < 0) {
    _b += 1000000007;
  }
  return ((long long int)((long long int)_a * (long long int)_b)) % 1000000007;
}
const int N = 2e5;
vector<long long int> vec, perm, new_vec;
long long int n, q, fac[20], tot, tree[N + 5];
void gen(long long int pos, long long int val, long long int mask) {
  if (pos == perm.size()) return;
  long long int len = (long long int)perm.size() - pos - 1, sum = 0, i, j;
  for (i = 0; i < perm.size(); ++i) {
    if (mask & 1ll << i) continue;
    if (tot >= sum + val + 1 && tot <= sum + val + fac[len]) {
      new_vec.push_back(perm[i]);
      gen(pos + 1, val + sum, mask | 1ll << i);
      return;
    }
    sum += fac[len];
  }
}
void update(long long int p, long long int v) {
  while (p <= n) {
    tree[p] += v;
    p += (p & (-p));
  }
}
long long int query(long long int p) {
  long long int ret = 0;
  while (p) {
    ret += tree[p];
    p ^= (p & (-p));
  }
  return ret;
}
long long int query(long long int l, long long int r) {
  if (l > r) return 0;
  return (query(r) - query(--l));
}
int main() {
  long long int i, j, k;
  scanf("%lld %lld", &n, &q);
  for (i = n, j = 0; i >= 1 && j < 14; ++j, --i) vec.push_back(i);
  for (i = 1, fac[0] = 1; i <= 15; ++i) fac[i] = fac[i - 1] * i;
  reverse(vec.begin(), vec.end());
  perm = vec;
  tot = 1;
  for (i = 1; i <= n; ++i) update(i, i);
  for (k = 0; k < q; ++k) {
    long long int t, l, r, x;
    scanf("%lld", &t);
    if (t == 1) {
      scanf("%lld %lld", &l, &r);
      long long int ret = query(l, r);
      printf("%lld", ret), puts("");
    } else if (t == 2) {
      scanf("%lld", &x);
      tot += x;
      new_vec.clear();
      gen(0, 0, 0);
      for (i = n, j = (long long int)vec.size() - 1; j >= 0; --j, --i) {
        update(i, -vec[j] + new_vec[j]);
      }
      vec = new_vec;
    }
  }
}