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)

#include <bits/stdc++.h>
#include <cxxabi.h>
using namespace std;
using ll=long long;
using vi=vector<int>;
using vvi=vector<vi>;
using pii=pair<int,int>;
#define rep(i,n) range(i,0,n)
#define range(i,a,n) for(int i=a;i<n;i++)
#define all(a) a.begin(),a.end()
#define LINF    ((ll)1ll<60)
#define INF     ((int)INT_MAX)
#define EPS     (1e-9)
#define MOD     (1000000007)
template<class S,class T>ostream& operator<<(ostream&os,pair<S,T>p){os<<"["<<p.first<<", "<<p.second<<"]";return os;};
template<class S>auto&operator<<(ostream&os,vector<S>t){bool b=1;for(auto s:t)os<<(exchange(b,0)||strlen(abi::__cxa_demangle(typeid(S).name(),0,0,0))>20?"":" ")<<s;return os<<endl;}

struct segTree{
	vi val;
	int N;
	segTree(int n){
		N=1;
		while(N<n)N*=2;
		val.resize(N*2);
		fill(all(val),INF);
	}
	
	void replace(int k, int a){
		k+=N;
		val[k]=a;
		while(k){
			k/=2;
			val[k]=min(val[k*2],val[k*2+1]);
		}
	}
	int minimum(int a, int b){return minimum(a,b,1,0,N);}
	int minimum(int a, int b, int k, int l, int r){
		if(r<=a or l>=b) return INF;
		if(a<=l and r<=b) return val[k];
		int m=(l+r)/2;
		return min(minimum(a,b,k*2,l,m),minimum(a,b,k*2+1,m,r));
	}	
};


int main(){
	cin.tie(0);
   	ios::sync_with_stdio(false);
	
	int n,q;
	while(cin>>n>>q){
		segTree st(n);
		rep(_,q){
			int c,x,y;
			cin>>c>>x>>y;
			if(c==0) st.replace(x,y);
			else cout<<st.minimum(x,y+1)<<endl;
		}
	}
	return 0;
}