#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pii pair<int, int>
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep3(i, l, n) for (int i = l; i < (n); ++i)
#define sz(v) (int)v.size()
const int inf = 1e9 + 7;
const ll INF = 1e18;
const int mod = 1000000007;
#define abs(x) (x >= 0 ? x : -(x))
#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())
#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }
template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<typename T> T pow(T a, int b) { return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1; }
ll modpow(ll a, int b, int _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }
template<class T> ostream& operator<<(ostream& os, const vector<T>& vec) { for (auto &vi: vec) os << vi << " "; return os; }
template<class T, class U> ostream& operator<<(ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }
template<typename T >inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }
template<class T> inline void add(T &a, int b) { a += b; if (a >= mod) a -= mod; }



void solve();

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(10);

    int T;
    // cin >> T;
    T = 1;

    while (T--) {
        solve();
    }
}

template<typename T>
class SegTree {
  int n;
  T def;
  vector<T> data;
  function<T(T, T)> operation; // query で
  function<T(T, T)> update; // change で

  T _query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return def;
    if (a <= l && r <= b) return data[k];
    T c1 = _query(a, b, 2 * k + 1, l, (l + r) / 2);
    T c2 = _query(a, b, 2 * k + 2, (l + r) / 2, r);
    return operation(c1, c2);
  }

public:
  SegTree<T>(int _n, T _def,
      function<T(T, T)> _operation, function<T(T, T)> _update)
    : def(_def), operation(_operation), update(_update) {
      n = 1;
      while (n < _n) n *= 2;
      data = vector<T>(2 * n - 1, def);
    }

  T query(int a, int b) {
    return _query(a, b, 0, 0, n);
  }

  void change(int i, T x) { // i を x に変える (operation は関係ない)
    i += n - 1;
    data[i] = update(data[i], x);
    while (i > 0) {
      i = (i - 1) / 2;
      data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
    }
  }

  T operator[](int i) { return data[i + n - 1]; }
  void out() { rep(i, n) cout << data[i + n - 1] << " "; cout << endl; }
};

void solve() {
  int n, q;
  cin >> n >> q;

  SegTree<int> seg(n, (1LL << 31) - 1,
    [](int a, int b) { return min(a, b); }, // query
    [](int a, int b) { return b; }); // change, ただ変更するだけ

  rep(i, q) {
    int com, x, y;
    cin >> com >> x >> y;
    if (com == 0) {
      seg.change(x, y);
      continue;
    }
    cout << seg.query(x, y + 1) << endl;
  }
}



Measured memory_bytes: 3820