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>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

typedef long long ll;
typedef pair<int, int> P;

const ll mod =  1000000007;  //出力は (ans % mod + mod) % mod  (負の剰余を正にする)
const int inf = 1e9;
const long long INF = 1LL << 60;   // INF = 1152921504606846976


const int MAX_N = 1 << 17;
ll n, dat[2*MAX_N-1];       //SegmentTreeを持つグローバル配列

void init(int n_){
    n = 1;
    while(n < n_) n *= 2;
    for(int i = 0; i < 2 * n - 1; i++) dat[i] = (1LL << 31)-1;
}

void update(ll k,ll a){
    // k番目の値をaに変更
    k += n-1;
    dat[k] = a;
    while(k > 0){
        k = (k-1)/2;
        dat[k] = min(dat[k*2 + 1],dat[k*2 + 2]);
    }
}

// [a,b)の最小値を求める.

ll query_sub(ll a, ll b, ll k, ll l, ll r){
    // k:現在見てるノードの位置, [l,r):dat[k] が表してる区間
    if(r <= a || b <= l){
        return  (1LL << 31)-1;
    }else if(a <= l && r <= b){
        //すべての区間がかぶる時
        return dat[k];
    }else{
        //一部の区間がかぶる時
        ll vl = query_sub(a,b,2*k + 1,l,(r+l)/2);
        ll vr = query_sub(a,b,2*k + 2,(l+r)/2,r);
        return min(vl,vr);
    }
}
ll query(ll a,ll b) {return query_sub(a, b, 0, 0, n);}

int main()
{
    int q;
    cin >> n >> q;
    init(n);
    rep(i,q){
        ll c,x,y;
        cin >> c >> x >> y;
        if(c == 0){
            update(x,y);
        }else{
            cout << query(x,y+1) << endl;;
        }
    }
}