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 std::max;
  using std::min;
  const int N = 200005;
  int n, q;
  int sum[N];
  long long sum1[N];
  inline void add(int p, int v) {
    while (p <= n) sum[p] += v, p += p & (-p);
  }
  inline int ask(int p) {
    int ret = 0;
    while (p) ret += sum[p], p -= p & (-p);
    return ret;
  }
  inline void add1(int p, int v) {
    while (p <= n) sum1[p] += v, p += p & (-p);
  }
  inline long long ask1(int p) {
    long long ret = 0;
    while (p) ret += sum1[p], p -= p & (-p);
    return ret;
  }
  inline void radd(int l, int r, int v) {
    add(l, v), add1(l, 1ll * l * v);
    add(r + 1, -v), add1(r + 1, -1ll * (r + 1) * v);
  }
  inline long long rask(int p) { return 1ll * (p + 1) * ask(p) - ask1(p); }
  namespace SGT {
  int mx[N << 2], mx1[N << 2], cnt[N << 2], tag[N << 2];
  inline void pushdown(int id) {
    mx[(id << 1)] = min(mx[(id << 1)], tag[id]);
    tag[(id << 1)] = min(tag[(id << 1)], tag[id]);
    mx[(id << 1 | 1)] = min(mx[(id << 1 | 1)], tag[id]);
    tag[(id << 1 | 1)] = min(tag[(id << 1 | 1)], tag[id]);
    tag[id] = 0x3f3f3f3f;
  }
  inline void upt(int id) {
    mx1[id] = 0;
    mx[id] = max(mx[(id << 1)], mx[(id << 1 | 1)]);
    cnt[id] = (mx[(id << 1)] == mx[id] ? cnt[(id << 1)] : 0) +
              (mx[(id << 1 | 1)] == mx[id] ? cnt[(id << 1 | 1)] : 0);
    mx1[id] = max(
        (mx[(id << 1)] == mx[id] ? mx1[(id << 1)] : mx[(id << 1)]),
        (mx[(id << 1 | 1)] == mx[id] ? mx1[(id << 1 | 1)] : mx[(id << 1 | 1)]));
  }
  void build(int id, int l, int r) {
    tag[id] = 0x3f3f3f3f;
    if (l == r) {
      mx[id] = l;
      cnt[id] = 1;
      return;
    }
    int mid = (l + r) >> 1;
    build((id << 1), l, mid), build((id << 1 | 1), mid + 1, r);
    upt(id);
  }
  void addp(int id, int l, int r, int p, int v) {
    if (l == r) {
      radd(l + 1, mx[id], -1);
      mx[id] = v;
      radd(l + 1, v, 1);
      return;
    }
    int mid = (l + r) >> 1;
    pushdown(id);
    if (p <= mid)
      addp((id << 1), l, mid, p, v);
    else
      addp((id << 1 | 1), mid + 1, r, p, v);
    upt(id);
  }
  void add(int id, int l, int r, int p) {
    if (mx[id] <= p) return;
    if (r <= p) {
      if (mx1[id] < p) {
        radd(p + 1, mx[id], -cnt[id]);
        tag[id] = p;
        mx[id] = p;
        return;
      }
    }
    pushdown(id);
    int mid = (l + r) >> 1;
    add((id << 1), l, mid, p);
    if (p > mid) add((id << 1 | 1), mid + 1, r, p);
    upt(id);
  }
  }  // namespace SGT
  using namespace SGT;
  int main() {
    scanf("%d%d", &n, &q);
    build(1, 1, n);
    for (int i = 1, tp, x, y; i <= q; ++i) {
      scanf("%d%d%d", &tp, &x, &y);
      if (tp == 1) {
        if (x - 1) add(1, 1, n, x - 1);
        addp(1, 1, n, x, y);
      } else
        printf("%lld\n", rask(y) - rask(x - 1) + y - x + 1);
    }
    return 0;
  }
  