description: Alice has recently received an array a_1, a_2, ..., a_n for her birthday! She is very proud of her array, and when she showed her friend Bob the array, he was very happy with her present too!
  
  However, soon Bob became curious, and as any sane friend would do, asked Alice to perform q operations of two types on her array:
  
    * 1 x y: update the element a_x to y (set a_x = y). 
    * 2 l r: calculate how many non-decreasing subarrays exist within the subarray [a_l, a_{l+1}, ..., a_r]. More formally, count the number of pairs of integers (p,q) such that l ≤ p ≤ q ≤ r and a_p ≤ a_{p+1} ≤ ... ≤ a_{q-1} ≤ a_q. 
  
  
  
  Help Alice answer Bob's queries!
  
  Input
  
  The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the size of the array, and the number of queries, respectively.
  
  The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the elements of Alice's array.
  
  The next q lines consist of three integers each. The first integer of the i-th line is t_i, the operation being performed on the i-th step (t_i = 1 or t_i = 2).
  
  If t_i = 1, the next two integers are x_i and y_i (1 ≤ x_i ≤ n; 1 ≤ y_i ≤ 10^9), updating the element at position x_i to y_i (setting a_{x_i} = y_i).
  
  If t_i = 2, the next two integers are l_i and r_i (1 ≤ l_i ≤ r_i ≤ n), the two indices Bob asks Alice about for the i-th query.
  
  It's guaranteed that there is at least one operation of the second type.
  
  Output
  
  For each query of type 2, print a single integer, the answer to the query.
  
  Example
  
  Input
  
  
  5 6
  3 1 4 1 5
  2 2 5
  2 1 3
  1 4 4
  2 2 5
  1 2 6
  2 2 5
  
  
  Output
  
  
  6
  4
  10
  7
  
  Note
  
  For the first query, l = 2 and r = 5, and the non-decreasing subarrays [p,q] are [2,2], [3,3], [4,4], [5,5], [2,3] and [4,5].
solution: #include <bits/stdc++.h>
  #pragma GCC target("avx2")
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  template <class T, class U>
  ostream &operator<<(ostream &os, const pair<T, U> &p) {
    os << '(' << p.first << ',' << p.second << ')';
    return os;
  }
  template <class T>
  ostream &operator<<(ostream &os, const vector<T> &v) {
    os << '{';
    for (int i = 0; i < (int)v.size(); i++) {
      if (i) {
        os << ',';
      }
      os << v[i];
    }
    os << '}';
    return os;
  }
  ostream &operator<<(ostream &os, const set<int> &se) {
    os << '{';
    int now = 0;
    for (auto x : se) {
      if (now) {
        os << ',';
      }
      os << x;
      now++;
    }
    os << '}';
    return os;
  }
  void debugg() { cerr << endl; }
  template <class T, class... Args>
  void debugg(const T &x, const Args &...args) {
    cerr << " " << x;
    debugg(args...);
  }
  using namespace std;
  const int inf = (1 << 30) - 1;
  const long long linf = 1LL << 61;
  const int MAX = 510000;
  int dy[8] = {0, 1, 0, -1, 1, -1, -1, 1};
  int dx[8] = {-1, 0, 1, 0, 1, -1, 1, -1};
  const double pi = acos(-1);
  const double eps = 1e-8;
  template <typename T1, typename T2>
  inline bool chmin(T1 &a, T2 b) {
    if (a > b) {
      a = b;
      return true;
    } else
      return false;
  }
  template <typename T1, typename T2>
  inline bool chmax(T1 &a, T2 b) {
    if (a < b) {
      a = b;
      return true;
    } else
      return false;
  }
  template <typename T>
  inline void print(T &a) {
    int sz = a.size();
    for (auto itr = a.begin(); itr != a.end(); itr++) {
      cout << *itr;
      sz--;
      if (sz) cout << " ";
    }
    cout << "\n";
  }
  template <typename T1, typename T2>
  inline void print2(T1 a, T2 b) {
    cout << a << " " << b << "\n";
  }
  template <typename T1, typename T2, typename T3>
  inline void print3(T1 a, T2 b, T3 c) {
    cout << a << " " << b << " " << c << "\n";
  }
  void mark() {
    cout << "#"
         << "\n";
  }
  long long pcount(long long x) { return __builtin_popcountll(x); }
  const int mod = 1e9 + 7;
  template <typename Monoid>
  struct Segtree {
    using F = function<Monoid(Monoid, Monoid)>;
    int sz;
    vector<Monoid> seg;
    const F f;
    const Monoid M;
    Segtree(int n, const F f, const Monoid &M) : f(f), M(M) {
      sz = 1;
      while (sz < n) sz <<= 1;
      seg.assign(2 * sz, M);
    }
    void set(int k, const Monoid &x) { seg[k + sz] = x; }
    void build() {
      for (int k = sz - 1; k > 0; k--) {
        seg[k] = f(seg[2 * k], seg[2 * k + 1]);
      }
    }
    void update(int k, const Monoid &x) {
      k += sz;
      seg[k] = x;
      while (k >>= 1) {
        seg[k] = f(seg[2 * k], seg[2 * k + 1]);
      }
    }
    Monoid query(int a, int b) {
      Monoid L = M, R = M;
      for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
        if (a & 1) L = f(L, seg[a++]);
        if (b & 1) R = f(seg[--b], R);
      }
      return f(L, R);
    }
    Monoid operator[](const int &k) const { return seg[k + sz]; }
    template <typename C>
    int find_subtree(int a, const C &check, Monoid &M0, bool type) {
      while (a < sz) {
        Monoid nxt = type ? f(seg[2 * a + type], M0) : f(M0, seg[2 * a + type]);
        if (check(nxt))
          a = 2 * a + type;
        else
          M0 = nxt, a = 2 * a + 1 - type;
      }
      return a - sz;
    }
    template <typename C>
    int find_first(int a, const C &check) {
      Monoid L = M;
      if (a <= 0) {
        if (check(f(L, seg[1]))) return find_subtree(1, check, L, false);
        return -1;
      }
      int b = sz;
      for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
        if (a & 1) {
          Monoid nxt = f(L, seg[a]);
          if (check(nxt)) return find_subtree(a, check, L, false);
          L = nxt;
          ++a;
        }
      }
      return -1;
    }
    template <typename C>
    int find_last(int b, const C &check) {
      Monoid R = M;
      if (b >= sz) {
        if (check(f(seg[1], R))) return find_subtree(1, check, R, true);
        return -1;
      }
      int a = sz;
      for (b += sz; a < b; a >>= 1, b >>= 1) {
        if (b & 1) {
          Monoid nxt = f(seg[--b], R);
          if (check(nxt)) return find_subtree(b, check, R, true);
          R = nxt;
        }
      }
      return -1;
    }
  };
  struct T {
    long long lval, rval, llen, rlen, sum, len;
  };
  T f(T a, T b) {
    T res;
    if (a.lval == -1) return b;
    if (b.lval == -1) return a;
    res.len = a.len + b.len;
    res.lval = a.lval;
    res.rval = b.rval;
    res.sum = a.sum + b.sum;
    if (a.rval <= b.lval) {
      res.sum -= a.rlen * (a.rlen + 1) / 2;
      res.sum -= b.llen * (b.llen + 1) / 2;
      long long L = a.rlen + b.llen;
      res.sum += L * (L + 1) / 2;
    }
    if (a.len == a.llen && b.len == b.rlen && a.rval <= b.lval) {
      res.llen = res.rlen = res.len;
    } else {
      if (a.len == a.llen && a.rval <= b.lval)
        res.llen = a.llen + b.llen;
      else
        res.llen = a.llen;
      if (b.len == b.rlen && a.rval <= b.lval)
        res.rlen = a.rlen + b.rlen;
      else
        res.rlen = b.rlen;
    }
    return res;
  }
  void solve() {
    int n, q;
    cin >> n >> q;
    vector<long long> a(n);
    for (int i = 0; i < (n); i++) cin >> a[i];
    Segtree<T> seg(n, f, T{-1, -1, 0, 0, 0, 0});
    for (int i = 0; i < (n); i++) seg.set(i, T{a[i], a[i], 1, 1, 1, 1});
    seg.build();
    while (q--) {
      int t, x, y;
      cin >> t >> x >> y;
      x--;
      if (t == 1) {
        seg.update(x, T{y, y, 1, 1, 1, 1});
      } else {
        cout << seg.query(x, y).sum << "\n";
      }
    }
  }
  int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int t = 1;
    while (t--) {
      solve();
    }
  }
  