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;
  const int MAXN = 200005;
  template <typename T>
  inline void read(T &WOW) {
    T x = 0, flag = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') flag = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    WOW = flag * x;
  }
  int n, q;
  struct BIT {
    long long t1[MAXN], t2[MAXN];
    inline void add(int x, int v) {
      long long tmp = x * v;
      for (int i = x; i <= n; i += (i & (-i))) {
        t1[i] += v;
        t2[i] += tmp;
      }
    }
    inline long long sum(int x) {
      long long r1 = 0, r2 = 0;
      for (int i = x; i; i -= (i & (-i))) {
        r1 += t1[i];
        r2 += t2[i];
      }
      return r1 * (x + 1) - r2;
    }
    inline void upd(int l, int r, int v) {
      add(l, v);
      if (r < n) add(r + 1, -v);
    }
    inline long long qry(int l, int r) { return sum(r) - sum(l - 1); }
  } seg1;
  struct SegmentTreeBeats {
    int mx[MAXN << 2], mx2[MAXN << 2], cnt[MAXN << 2], tag[MAXN << 2];
    inline void puttag(int rt, int tg) { mx[rt] = tag[rt] = tg; }
    inline void pushdown(int rt) {
      if (tag[rt]) {
        int tmp = max(mx[(rt << 1)], mx[(rt << 1) | 1]);
        if (mx[(rt << 1)] == tmp) {
          puttag((rt << 1), tag[rt]);
        }
        if (mx[(rt << 1) | 1] == tmp) {
          puttag((rt << 1) | 1, tag[rt]);
        }
        tag[rt] = 0;
      }
    }
    inline void pushup(int rt) {
      if (mx[(rt << 1)] == mx[(rt << 1) | 1]) {
        mx[rt] = mx[(rt << 1)];
        mx2[rt] = max(mx2[(rt << 1)], mx2[(rt << 1) | 1]);
        cnt[rt] = cnt[(rt << 1)] + cnt[(rt << 1) | 1];
      } else if (mx[(rt << 1)] > mx[(rt << 1) | 1]) {
        mx[rt] = mx[(rt << 1)];
        mx2[rt] = max(mx2[(rt << 1)], mx[(rt << 1) | 1]);
        cnt[rt] = cnt[(rt << 1)];
      } else {
        mx[rt] = mx[(rt << 1) | 1];
        mx2[rt] = max(mx[(rt << 1)], mx2[(rt << 1) | 1]);
        cnt[rt] = cnt[(rt << 1) | 1];
      }
    }
    void Build(int rt, int b, int e) {
      if (b == e) {
        mx[rt] = b;
        mx2[rt] = 0;
        cnt[rt] = 1;
        return;
      }
      int mid = (b + e) >> 1;
      Build((rt << 1), b, mid);
      Build((rt << 1) | 1, mid + 1, e);
      pushup(rt);
    }
    void Insert(int rt, int b, int e, int p, int v) {
      if (b == e) {
        seg1.upd(b, mx[rt], -1);
        mx[rt] = v;
        seg1.upd(b, mx[rt], 1);
        return;
      }
      int mid = (b + e) >> 1;
      pushdown(rt);
      if (p <= mid)
        Insert((rt << 1), b, mid, p, v);
      else
        Insert((rt << 1) | 1, mid + 1, e, p, v);
      pushup(rt);
    }
    void UpdateMin(int rt, int b, int e, int l, int r, int v) {
      if (mx[rt] <= v) return;
      if (l <= b && e <= r && v > mx2[rt]) {
        seg1.upd(v + 1, mx[rt], -cnt[rt]);
        puttag(rt, v);
        return;
      }
      int mid = (b + e) >> 1;
      pushdown(rt);
      if (l <= mid) UpdateMin((rt << 1), b, mid, l, r, v);
      if (r > mid) UpdateMin((rt << 1) | 1, mid + 1, e, l, r, v);
      pushup(rt);
    }
  } seg2;
  void solve() {
    read(n);
    read(q);
    seg1.upd(1, n, 1);
    seg2.Build(1, 1, n);
    int type, l, r;
    while (q--) {
      read(type);
      read(l);
      read(r);
      if (type == 1) {
        seg2.Insert(1, 1, n, l, r);
        if (l > 1) seg2.UpdateMin(1, 1, n, 1, l - 1, l - 1);
      } else {
        printf("%lld\n", seg1.qry(l, r));
      }
    }
  }
  int main() {
    solve();
    return 0;
  }
  