Problem:
Range Minimum Query (RMQ)

Write a program which manipulates a sequence A = {a0, a1, . . . , an-1} with the following operations:

find(s, t): report the minimum element in as, as+1, . . . ,at.

update(i, x): change ai to x.

Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.

Input

n q
com0 x0 y0
com1 x1 y1
...
comq−1 xq−1 yq−1

In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes update(xi, yi) and '1' denotes find(xi, yi).

Output

For each find operation, print the minimum element.

Constraints

1 ≤ n ≤ 100000

1 ≤ q ≤ 100000

If comi is 0, then 0 ≤ xi < n, 0 ≤ yi < 231-1.

If comi is 1, then 0 ≤ xi < n, 0 ≤ yi < n.

Sample Input 1

3 5
0 0 1
0 1 2
0 2 3
1 0 2
1 1 2

Sample Output 1

1
2

Sample Input 2

1 3
1 0 0
0 0 5
1 0 0

Sample Output 2

2147483647
5

Solution in C++14 (judge verdict: Accepted)

#pragma GCC optimize("O3")
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
#include <bitset>
using namespace std;

using ll = long long;
using P = pair<int, int>;
using T = tuple<int, int, int>;

template <class T> inline T chmax(T &a, const T b) {return a = (a < b) ? b : a;}
template <class T> inline T chmin(T &a, const T b) {return a = (a > b) ? b : a;}

constexpr int MOD = 1e9 + 7;
constexpr int inf = 1e9;
constexpr long long INF = 1e18;
constexpr double pi = acos(-1);
constexpr double EPS = 1e-10;

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

class SegmentTree{
public:
    int n;
    vector<ll> node;

    // Range Minumum Query
    SegmentTree(vector<ll> v){
        int sz = v.size();
        n = 1; while(n < sz) n *= 2;
        node.resize(2*n-1, INF);
        for(int i=0; i<sz; i++) node[i+n-1] = v[i];
        for(int i=n-2; i>=0; i--) node[i] = min(node[2*i+1], node[2*i+2]);
    }

    void update(int x, ll val){
        x += (n - 1);
        node[x] = val;
        while(0 < x){
            x = (x - 1) / 2;
            node[x] = min(node[2*x+1], node[2*x+2]);
        }
    }

    ll getmin(int a, int b, int k=0, int l=0, int r=-1){
        if(r < 0) r = n;
        if(r <= a || b <= l) return INF;
        if(a <= l && r <= b) return node[k];

        ll vl = getmin(a, b, 2*k+1, l, (l+r)/2);
        ll vr = getmin(a, b, 2*k+2, (l+r)/2, r);
        
        return min(vl, vr);
    }
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, q; cin>>n>>q;
    vector<ll> a(n, (1LL<<31)-1);
    SegmentTree seg(a);

    while(q--){
        int c, x, y; cin>>c>>x>>y;
        if(c == 0) seg.update(x, y);
        else cout << seg.getmin(x, y + 1) << endl;
    }

    return 0;
}