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;
  inline long long readint() {
    long long x = 0;
    bool f = 0;
    char c = getchar();
    while (!isdigit(c) && c != '-') c = getchar();
    if (c == '-') {
      f = 1;
      c = getchar();
    }
    while (isdigit(c)) {
      x = x * 10 + c - '0';
      c = getchar();
    }
    return f ? -x : x;
  }
  int n, q;
  struct node1 {
    int l, r;
    node1* ch[2];
    long long sum, add = 0;
    void pushup() { sum = ch[0]->sum + ch[1]->sum; }
    node1(int l, int r) : l(l), r(r), sum(r - l + 1) {
      if (l == r) return;
      int mid = l + (r - l) / 2;
      ch[0] = new node1(l, mid);
      ch[1] = new node1(mid + 1, r);
    }
    void pushtag(long long k) {
      sum += k * (r - l + 1);
      add += k;
    }
    void pushdown() {
      ch[0]->pushtag(add);
      ch[1]->pushtag(add);
      add = 0;
    }
    void modify(int ql, int qr, long long k) {
      if (ql <= l && qr >= r) {
        pushtag(k);
        return;
      }
      pushdown();
      if (ql <= ch[0]->r) ch[0]->modify(ql, qr, k);
      if (qr >= ch[1]->l) ch[1]->modify(ql, qr, k);
      pushup();
    }
    long long query(int ql, int qr) {
      if (ql <= l && qr >= r) return sum;
      pushdown();
      long long ans = 0;
      if (ql <= ch[0]->r) ans += ch[0]->query(ql, qr);
      if (qr >= ch[1]->l) ans += ch[1]->query(ql, qr);
      return ans;
    }
  } * rt1;
  struct node2 {
    int l, r;
    node2* ch[2];
    int mx, mx2, cnt = 1, add = 0;
    void pushup() {
      if (ch[0]->mx > ch[1]->mx) {
        mx = ch[0]->mx;
        mx2 = max(ch[0]->mx2, ch[1]->mx);
        cnt = ch[0]->cnt;
      } else if (ch[0]->mx < ch[1]->mx) {
        mx = ch[1]->mx;
        mx2 = max(ch[1]->mx2, ch[0]->mx);
        cnt = ch[1]->cnt;
      } else {
        mx = ch[0]->mx;
        mx2 = max(ch[0]->mx2, ch[1]->mx2);
        cnt = ch[0]->cnt + ch[1]->cnt;
      }
    }
    node2(int l, int r) : l(l), r(r), mx(r), mx2(l == r ? 0 : r - 1) {
      if (l == r) return;
      int mid = l + (r - l) / 2;
      ch[0] = new node2(l, mid);
      ch[1] = new node2(mid + 1, r);
    }
    void pushtag(int k) {
      mx += k;
      add += k;
    }
    void pushdown() {
      if (ch[0]->mx > ch[1]->mx)
        ch[0]->pushtag(add);
      else if (ch[0]->mx < ch[1]->mx)
        ch[1]->pushtag(add);
      else {
        ch[0]->pushtag(add);
        ch[1]->pushtag(add);
      }
      add = 0;
    }
    void modify1(int x, int k) {
      if (l == r) {
        rt1->modify(x, mx, -1);
        rt1->modify(x, mx = k, 1);
        return;
      }
      pushdown();
      if (x <= ch[0]->r)
        ch[0]->modify1(x, k);
      else
        ch[1]->modify1(x, k);
      pushup();
    }
    void modify2(int ql, int qr, int k) {
      if (k >= mx) return;
      if (ql <= l && qr >= r && k >= mx2) {
        rt1->modify(k + 1, mx, -cnt);
        pushtag(k - mx);
        return;
      }
      pushdown();
      if (ql <= ch[0]->r) ch[0]->modify2(ql, qr, k);
      if (qr >= ch[1]->l) ch[1]->modify2(ql, qr, k);
      pushup();
    }
  } * rt2;
  int main() {
    n = readint();
    q = readint();
    rt1 = new node1(1, n);
    rt2 = new node2(1, n);
    while (q--) {
      int p = readint();
      if (p == 1) {
        int c, g;
        c = readint();
        g = readint();
        rt2->modify1(c, g);
        if (c > 1) rt2->modify2(1, c - 1, c - 1);
      } else {
        int l, r;
        l = readint();
        r = readint();
        printf("%lld\n", rt1->query(l, r));
      }
    }
    return 0;
  }
  