description: There are n cities in a row numbered from 1 to n.
  
  The cities will be building broadcasting stations. The station in the i-th city has height h_i and range w_i. It can broadcast information to city j if the following constraints are met: 
  
    * i ≤ j ≤ w_i, and 
    * for each k such that i < k ≤ j, the following condition holds: h_k < h_i. 
  
  In other words, the station in city i can broadcast information to city j if j ≥ i, j is in the range of i-th station, and i is strictly highest on the range from i to j (including city j).
  
  At the beginning, for every city i, h_i = 0 and w_i = i.
  
  Then q events will take place. During i-th event one of the following will happen: 
  
    * City c_i will rebuild its station so that its height will be strictly highest among all stations and w_{c_i} will be set to g_i. 
    * Let b_j be the number of stations that can broadcast information to city j. Print the sum of b_j over all j satisfying l_i ≤ j ≤ r_i. 
  
  
  
  Your task is to react to all events and print answers to all queries.
  
  Input
  
  The first line contains two integers n and q (1 ≤ n, q ≤ 2⋅10^5) — number of cities and number of events.
  
  Then q lines follow. The i-th line begins with an integer p_i (p_i = 1 or p_i = 2).
  
  If p_i = 1 a station will be rebuilt. Then two integers c_i and g_i (1 ≤ c_i ≤ g_i ≤ n) follow — the city in which the station is rebuilt and its new broadcasting range.
  
  If p_i = 2 you are given a query. Then two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) follow — the range of cities in the query.
  
  Output
  
  For each query, print in a single line the sum of b_j over the given interval.
  
  Examples
  
  Input
  
  
  1 3
  2 1 1
  1 1 1
  2 1 1
  
  
  Output
  
  
  1
  1
  
  
  Input
  
  
  5 10
  1 3 4
  2 3 5
  1 1 5
  2 1 5
  1 4 5
  2 2 4
  1 2 3
  2 1 3
  1 5 5
  2 2 5
  
  
  Output
  
  
  4
  10
  5
  4
  5
  
  Note
  
  In the first test case, only station 1 reaches city 1 before and after it is rebuilt.
  
  In the second test case, after each rebuild, the array b looks as follows: 
  
    1. [1, 1, 1, 2, 1]; 
    2. [1, 2, 2, 3, 2]; 
    3. [1, 2, 2, 1, 2]; 
    4. [1, 1, 2, 1, 2]; 
    5. [1, 1, 2, 1, 1]. 
solution: #include <bits/stdc++.h>
  using namespace std;
  int n, m;
  struct BIT {
    long long t1[200005], t2[200005];
    void mdf(int p, int v) {
      for (long long x = 1ll * p * v; p <= n; p += (p & -p))
        t1[p] += v, t2[p] += x;
    }
    void mdf(int l, int r, int v) { mdf(l, v), mdf(r + 1, -v); }
    long long _q(int x, long long r1 = 0, long long r2 = 0) {
      for (int p = x; p; p -= (p & -p)) r1 += t1[p], r2 += t2[p];
      return r1 * (x + 1) - r2;
    }
    long long qry(int l, int r) { return _q(r) - _q(l - 1); }
  } B;
  int mx[200005 << 2], sc[200005 << 2], cnt[200005 << 2], tg[200005 << 2];
  void cov(int k, int v) {
    if (v > mx[k]) return;
    mx[k] = tg[k] = v;
  }
  void pud(int k) {
    if (~tg[k]) cov(k << 1, tg[k]), cov(k << 1 | 1, tg[k]), tg[k] = -1;
  }
  void upd(int k) {
    int ls = k << 1, rs = k << 1 | 1;
    mx[k] = max(mx[ls], mx[rs]);
    cnt[k] = 0;
    sc[k] = -1;
    if (mx[k] == mx[ls])
      cnt[k] += cnt[ls], sc[k] = max(sc[k], sc[ls]);
    else
      sc[k] = max(sc[k], mx[ls]);
    if (mx[k] == mx[rs])
      cnt[k] += cnt[rs], sc[k] = max(sc[k], sc[rs]);
    else
      sc[k] = max(sc[k], mx[rs]);
  }
  void build(int k, int l, int r) {
    tg[k] = -1;
    if (l == r) {
      mx[k] = l, sc[k] = -1, cnt[k] = 1, B.mdf(l, l, 1);
      return;
    }
    int mid = l + r >> 1;
    build(k << 1, l, mid), build(k << 1 | 1, mid + 1, r);
    upd(k);
  }
  void Min(int k, int l, int r, int x, int y, int v) {
    if (v >= mx[k]) return;
    if (x <= l && r <= y && v > sc[k])
      return B.mdf(v + 1, mx[k], -cnt[k]), cov(k, v);
    pud(k);
    int mid = l + r >> 1;
    if (x <= mid) Min(k << 1, l, mid, x, y, v);
    if (y > mid) Min(k << 1 | 1, mid + 1, r, x, y, v);
    upd(k);
  }
  void mdf(int k, int l, int r, int x, int v) {
    if (l == r) {
      B.mdf(l, mx[k], -1), mx[k] = v, tg[k] = -1, B.mdf(l, v, 1);
      return;
    }
    pud(k);
    int mid = l + r >> 1;
    (x <= mid) ? mdf(k << 1, l, mid, x, v) : mdf(k << 1 | 1, mid + 1, r, x, v);
    upd(k);
  }
  int main() {
    scanf("%d%d", &n, &m);
    build(1, 1, n);
    for (int opt, x, y; m--;) {
      scanf("%d%d%d", &opt, &x, &y);
      if (opt == 1) {
        if (x > 1) Min(1, 1, n, 1, x - 1, x - 1);
        mdf(1, 1, n, x, y);
      } else
        printf("%lld\n", B.qry(x, y));
    }
  }
  