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>
  void sc(int &x) { scanf("%d", &x); }
  void sc(int &x, int &y) { scanf("%d%d", &x, &y); }
  void sc(int &x, int &y, int &z) { scanf("%d%d%d", &x, &y, &z); }
  void sc(long long &x) { scanf("%lld", &x); }
  void sc(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
  void sc(long long &x, long long &y, long long &z) {
    scanf("%lld%lld%lld", &x, &y, &z);
  }
  void sc(char *x) { scanf("%s", x); }
  void sc(char *x, char *y) { scanf("%s%s", x, y); }
  void sc(char *x, char *y, char *z) { scanf("%s%s%s", x, y, z); }
  void out(int x) { printf("%d\n", x); }
  void out(long long x) { printf("%lld\n", x); }
  void out(int x, int y) { printf("%d %d\n", x, y); }
  void out(long long x, long long y) { printf("%lld %lld\n", x, y); }
  void out(int x, int y, int z) { printf("%d %d %d\n", x, y, z); }
  void out(long long x, long long y, long long z) {
    printf("%lld %lld %lld\n", x, y, z);
  }
  void ast(long long x, long long l, long long r) { assert(x >= l && x <= r); }
  using namespace std;
  const int N = 3e5 + 5, mod = 998244353;
  int n, q;
  long long b[N << 2], lz[N << 2];
  void up(int l, int r, int k) {
    b[k] += 1ll * lz[k] * (r - l + 1);
    if (l != r) lz[k << 1] += lz[k], lz[k << 1 | 1] += lz[k];
    lz[k] = 0;
  }
  void ins(int l, int r, int k, int x, int y, int v) {
    if (lz[k]) up(l, r, k);
    if (r < x || l > y) return;
    if (l >= x && r <= y) {
      lz[k] += v;
      up(l, r, k);
      return;
    }
    int m = (l + r) >> 1;
    ins(l, m, k << 1, x, y, v);
    ins(m + 1, r, k << 1 | 1, x, y, v);
    b[k] = b[k << 1] + b[k << 1 | 1];
  }
  long long query(int l, int r, int k, int x, int y) {
    if (lz[k]) up(l, r, k);
    if (r < x || l > y) return 0;
    if (l >= x && r <= y) return b[k];
    int m = (l + r) >> 1;
    return query(l, m, k << 1, x, y) + query(m + 1, r, k << 1 | 1, x, y);
  }
  struct node {
    int mx, se, mxnum, lz;
    node operator+(const node &o) const {
      node ans;
      ans.se = max(se, o.se);
      if (mx == o.mx)
        ans.mx = mx, ans.mxnum = mxnum + o.mxnum;
      else if (mx > o.mx)
        ans.mx = mx, ans.mxnum = mxnum, ans.se = max(ans.se, o.mx);
      else
        ans.mx = o.mx, ans.mxnum = o.mxnum, ans.se = max(ans.se, mx);
      ans.lz = mod;
      return ans;
    }
  } t[N << 2];
  bool debug = false;
  void build(int l, int r, int k) {
    if (l == r) {
      t[k].mx = l;
      t[k].mxnum = 1;
      t[k].se = -1;
      t[k].lz = mod;
      return;
    }
    int m = (l + r) >> 1;
    build(l, m, k << 1);
    build(m + 1, r, k << 1 | 1);
    t[k] = t[k << 1] + t[k << 1 | 1];
  }
  void update(int l, int r, int k) {
    if (t[k].lz >= t[k].mx) {
      t[k].lz = mod;
      return;
    }
    t[k].mx = t[k].lz;
    if (l != r)
      t[k << 1].lz = min(t[k << 1].lz, t[k].lz),
             t[k << 1 | 1].lz = min(t[k << 1 | 1].lz, t[k].lz);
    t[k].lz = mod;
  }
  void setv(int l, int r, int k, int x, int v) {
    if (t[k].lz != mod) update(l, r, k);
    if (l == r) {
      if (v > t[k].mx) {
        if (debug) cout << "ins: " << t[k].mx + 1 << ' ' << v << ' ' << 1 << '\n';
        ins(1, n, 1, t[k].mx + 1, v, 1);
      } else if (v < t[k].mx) {
        if (debug)
          cout << "ins: " << v + 1 << ' ' << t[k].mx << ' ' << -1 << '\n';
        ins(1, n, 1, v + 1, t[k].mx, -1);
      }
      t[k].mx = v;
      return;
    }
    int m = (l + r) >> 1;
    if (x <= m)
      setv(l, m, k << 1, x, v), update(m + 1, r, k << 1 | 1);
    else
      setv(m + 1, r, k << 1 | 1, x, v), update(l, m, k << 1);
    t[k] = t[k << 1] + t[k << 1 | 1];
  }
  void setmn(int l, int r, int k, int x, int y, int v) {
    if (t[k].lz != mod) update(l, r, k);
    if (r < x || l > y) return;
    if (l >= x && r <= y && v > t[k].se) {
      if (debug) cout << "ok " << l << ' ' << r << ' ' << t[k].mx << '\n';
      if (v < t[k].mx) {
        if (debug)
          cout << "ins: " << v + 1 << ' ' << t[k].mx << ' ' << -t[k].mxnum
               << '\n';
        ins(1, n, 1, v + 1, t[k].mx, -t[k].mxnum);
      }
      t[k].lz = v;
      update(l, r, k);
      return;
    }
    int m = (l + r) >> 1;
    setmn(l, m, k << 1, x, y, v);
    setmn(m + 1, r, k << 1 | 1, x, y, v);
    t[k] = t[k << 1] + t[k << 1 | 1];
  }
  void sol(int cas) {
    sc(n, q);
    build(1, n, 1);
    while (q--) {
      int op, x, y;
      sc(op, x, y);
      if (op == 2)
        printf("%lld\n", query(1, n, 1, x, y) + y - x + 1);
      else {
        setv(1, n, 1, x, y);
        if (x > 1) setmn(1, n, 1, 1, x - 1, x - 1);
      }
    }
  }
  int main() {
    srand(time(0));
    int t = 1, cas = 0;
    while (t--) {
      sol(++cas);
    }
    return 0;
  }
  