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++11 (judge verdict: Time Limit Exceeded)

/*??????????????§RMQ????±???????*/
#include <iostream>
#include <vector>
#include <math.h>
using namespace std;
#define RMQ_MAX ((1LL << 31) - 1)
template<typename T>
class SquareBucket{
public:
  SquareBucket(){}
  SquareBucket(const vector<T> &array):val(array){
    N = val.size();
    sqrtN = ceil(sqrt(N));
    B = (N + sqrtN - 1) / sqrtN;
    bucket.assign(B, RMQ_MAX);
    for (int i = 0; i < B; i++){
      for (int j = sqrtN * i; j < sqrtN * (i + 1); j++){
	if (j >= N)break;
	bucket[i] = min(bucket[i], val[j]);
      }
    }
  }
  void init(const vector<T> &array){
    val = array;
    N = val.size();
    sqrtN = ceil(sqrt(N));
    B = (N + sqrtN - 1) / sqrtN;
    bucket.assign(B, RMQ_MAX);
    for (int i = 0; i < B; i++){
      for (int j = sqrtN * i; j < sqrtN * (i + 1); j++){
	if (j >= N)break;
	bucket[i] = min(bucket[i], val[j]);
      }
    }

  }
  /*[l, r)???RMQ*/
  T range_minimum(int left, int right){
    int L = (left + sqrtN - 1) / sqrtN + 1;
    int R = right / sqrtN;
    T res = RMQ_MAX;
    if (L <= R){
      for (int i = left; i < sqrtN * L; i++){
	res = min(res, val[i]);
      }
      for (int i = L; i < R; i++){
	res = min(res, bucket[i]);
      }
      for (int i = sqrtN * R; i < right; i++){
	res = min(res, val[i]);
      }
    }else{
      for (int i = left; i < right; i++){
	res = min(res, val[i]);
      }
    }
    return res;
  }
private:
  vector<T> val;
  vector<T> bucket;
  int N;
  int sqrtN;
  int B;//the number of bucket
};

int N,Q;
int main(){
  cin >> N >> Q;
  vector<int> arr;
  arr.resize(N, RMQ_MAX);
  for (int i = 0; i < Q; i++){
    int com,x,y;
    cin >>com>> x >> y;
    if (com == 0){
      arr[x] = y;
    }else{
      SquareBucket<int> rmq(arr);
      cout << rmq.range_minimum(x, y + 1) << endl;
    }
  }
}