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++ (judge verdict: Accepted)

#include<stdio.h>
#include<iostream>
#define VALUE 2147483647

using namespace std;
int main() {
	int n, q;
	cin >> n >> q;

	int a[100000];
	//初期値の設定
	for (int i = 0; i < n; i++) {
		a[i] = VALUE;
	}

	int com, x, y;
	for (int i = 0; i < q; i++) {
		cin >> com >> x >> y;

		if (com == 0) {
			a[x] = y;
		}
		else {
			int mini = VALUE;
			int j = x;
			while(1){
				
				if (a[j] < mini) {
					mini = a[j];
				}
				if (j == y) break;

				j++;
			}
			cout << mini << endl;
		}
	}

	return 0;
}