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>
  #pragma GCC optimize(2, 3, "Ofast")
  #pragma GCC target("avx", "avx2")
  using namespace std;
  template <typename T1, typename T2>
  void ckmin(T1 &a, T2 b) {
    if (a > b) a = b;
  }
  template <typename T1, typename T2>
  void ckmax(T1 &a, T2 b) {
    if (a < b) a = b;
  }
  int read() {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= ch == '-', ch = getchar();
    while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
    return f ? -x : x;
  }
  template <typename T>
  void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x / 10);
    putchar(x % 10 + '0');
  }
  template <typename T>
  void print(T x, char let) {
    print(x), putchar(let);
  }
  const int N = 200005;
  const int inf = 1e9;
  int n, q;
  struct SMT1 {
    long long sum[N << 2], tag[N << 2];
    void pushup(int u) { sum[u] = sum[(u << 1)] + sum[(u << 1 | 1)]; }
    void pushtag(int u, int tg, int l, int r) {
      tag[u] += tg, sum[u] += 1ll * tg * (r - l + 1);
    }
    void pushdown(int u, int l, int r) {
      if (tag[u]) {
        int mid = l + r >> 1;
        pushtag((u << 1), tag[u], l, mid),
            pushtag((u << 1 | 1), tag[u], mid + 1, r);
        tag[u] = 0;
      }
    }
    void build(int u, int l, int r) {
      if (l == r) {
        sum[u] = 1;
        return;
      }
      int mid = l + r >> 1;
      build((u << 1), l, mid), build((u << 1 | 1), mid + 1, r);
      pushup(u);
    }
    void update(int u, int l, int r, int ql, int qr, int x) {
      if (ql > qr) return;
      if (ql <= l && r <= qr) {
        pushtag(u, x, l, r);
        return;
      }
      pushdown(u, l, r);
      int mid = l + r >> 1;
      if (ql <= mid) update((u << 1), l, mid, ql, qr, x);
      if (qr > mid) update((u << 1 | 1), mid + 1, r, ql, qr, x);
      pushup(u);
    }
    long long query(int u, int l, int r, int ql, int qr) {
      if (ql <= l && r <= qr) return sum[u];
      int mid = l + r >> 1;
      pushdown(u, l, r);
      long long ret = 0;
      if (ql <= mid) ret += query((u << 1), l, mid, ql, qr);
      if (qr > mid) ret += query((u << 1 | 1), mid + 1, r, ql, qr);
      return ret;
    }
  } smt1;
  struct SMT2 {
    int mx[N << 2], mx2[N << 2], cnt[N << 2];
    int tag[N << 2];
    void pushup(int u) {
      if (mx[(u << 1)] == mx[(u << 1 | 1)])
        mx[u] = mx[(u << 1)], mx2[u] = max(mx2[(u << 1)], mx2[(u << 1 | 1)]),
        cnt[u] = cnt[(u << 1)] + cnt[(u << 1 | 1)];
      else if (mx[(u << 1)] > mx[(u << 1 | 1)])
        mx[u] = mx[(u << 1)], mx2[u] = max(mx2[(u << 1)], mx[(u << 1 | 1)]),
        cnt[u] = cnt[(u << 1)];
      else
        mx[u] = mx[(u << 1 | 1)], mx2[u] = max(mx[(u << 1)], mx2[(u << 1 | 1)]),
        cnt[u] = cnt[(u << 1 | 1)];
    }
    void pushtag(int u, int t) {
      if (mx[u] > t) mx[u] = t, tag[u] = t;
    }
    void pushdown(int u) {
      if (tag[u] != inf)
        pushtag((u << 1), tag[u]), pushtag((u << 1 | 1), tag[u]), tag[u] = inf;
    }
    void build(int u, int l, int r) {
      tag[u] = inf;
      if (l == r) {
        mx[u] = l, mx2[u] = 0, cnt[u] = 1;
        return;
      }
      int mid = l + r >> 1;
      build((u << 1), l, mid), build((u << 1 | 1), mid + 1, r);
      pushup(u);
    }
    void beat(int u, int l, int r, int ql, int qr, int x) {
      if (ql > qr || mx[u] <= x) return;
      if (ql <= l && r <= qr && mx2[u] < x) {
        smt1.update(1, 1, n, x + 1, mx[u], -cnt[u]);
        pushtag(u, x);
        return;
      }
      pushdown(u);
      int mid = l + r >> 1;
      if (ql <= mid) beat((u << 1), l, mid, ql, qr, x);
      if (qr > mid) beat((u << 1 | 1), mid + 1, r, ql, qr, x);
      pushup(u);
    }
    void update(int u, int l, int r, int pos, int x) {
      if (l == r) {
        if (x > mx[u])
          smt1.update(1, 1, n, mx[u] + 1, x, 1);
        else
          smt1.update(1, 1, n, x + 1, mx[u], -1);
        mx[u] = x, mx2[u] = 0, cnt[u] = 1;
        return;
      }
      pushdown(u);
      int mid = l + r >> 1;
      if (pos <= mid)
        update((u << 1), l, mid, pos, x);
      else
        update((u << 1 | 1), mid + 1, r, pos, x);
      pushup(u);
    }
  } smt2;
  int main() {
    n = read(), q = read();
    smt1.build(1, 1, n);
    smt2.build(1, 1, n);
    while (q--) {
      int opt = read(), x = read(), y = read();
      if (opt == 1) {
        smt2.beat(1, 1, n, 1, x - 1, x - 1);
        smt2.update(1, 1, n, x, y);
      } else {
        print(smt1.query(1, 1, n, x, y), '\n');
      }
    }
    return 0;
  }
  