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: Time Limit Exceeded)

#include <stdio.h>

#define RANGE_MAX 2147483647
#define MAX 100005

int A[MAX];

int main(void) {
	int i, j, n, q, m;
	int com, x, y, min;
	scanf("%d %d\n", &n, &q);
	for(i=0;i<n;i++) A[i] = RANGE_MAX;
	
	for(i=0;i<q;i++) {
		scanf("%d %d %d", &com, &x, &y);
		if(com == 0) {
			A[x] = y;
		} else { 
			m = RANGE_MAX;
			for(j=x;j<=y;j++) {
				if(m > A[j])
					m = A[j];
			}
			printf("%d\n", m);
		}
	}
	
	return 0;
}