// Original C++ solution (slower):
#include<iostream>

#include<vector>

#include<algorithm>

#include<queue>

#include<map>

#include<stack>

#include<cmath>

#include<iomanip>

#include<set>

#include<numeric>

#include<sstream>

#include<random>

#include<cassert>

#include<climits>

using namespace std;

typedef long long ll;

#define rep(i, n) for (int i = 0; i < n; ++i)

#define rrep(i, st, n) for (int i = st; i < n; ++i)

using pii = pair<int, int>;

const int inf = 1e9 + 7;

int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};

int dx[] = {1, -1, 0, 0, -1, 1, 1, -1};

#define ceil(a, b) a / b + !!(a % b)

#define chmax(a, b) a = max(a, b)

#define chmin(a, b) a = min(a, b)



const int MAX_N = 131072;

int n;

int dat[MAX_N * 2 - 1];

void init() {

    rep(i, 2 * MAX_N - 1) dat[i] = INT_MAX;

}

void update(int i, int x) {

    //高さ4の木

    //i = 2なら木ではindex 9

    i += n - 1;

    dat[i] = x;

    //cout << i << " " << x << endl;

    while (i > 0) {

        i = (i - 1) / 2; //登る

        dat[i] = min(dat[i * 2 + 1], dat[i * 2 + 2]);

    }

}

int query(int a, int b, int k, int l, int r) {

    if (r <= a || b <= l) return INT_MAX;

    if (a <= l && r <= b) return dat[k];

    int vl = query(a, b, k * 2 + 1, l, (l + r) / 2);

    int vr = query(a, b, k * 2 + 2, (l + r) / 2, r);

    return min(vl, vr);

    

}

int main() {

    int q; cin >> n >> q;

    int tmp = 1; while (tmp < n) tmp *= 2;

    n = tmp;

    init(); //cin >> nのあとじゃないとsize?になるかも

    rep(i, q) {

        int com, x, y; cin >> com >> x >> y;

        if (com == 0) {

            update(x, y);

        } else {

            cout << query(x, y + 1, 0, 0, n) << endl;

        }

        //cout << endl;

        //rep(j, 2 * n + 1) cout << dat[j] << " "; cout << endl;

    }

}




// Optimized solution to the same problem (faster):
#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;

  }

}




// Fastest known solution for this problem: