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: Accepted)

#include <iostream>
#include <algorithm>
using namespace std;
int n, q, c, x, y, size_, dat[266666];
inline void update(int pos, int x) {
	pos += size_; dat[pos] = x;
	while(pos != 1) {
		pos >>= 1;
		dat[pos] = min(dat[pos * 2], dat[pos * 2 + 1]);
	}
}
inline int query(int a, int b, int k, int l, int r) {
	if(r <= a || b <= l) return 2147483647;
	if(a <= l && r <= b) return dat[k]; k <<= 1;
	int lc = query(a, b, k, l, (l + r) / 2);
	int rc = query(a, b, k + 1, (l + r) / 2, r);
	return min(lc, rc);
}
int main() {
	scanf("%d%d", &n, &q);
	for(size_ = 1; size_ < n;) size_ *= 2;
	for(int i = 0; i < size_ * 2; i++) dat[i] = 2147483647;
	for(int i = 0; i < q; i++) {
		scanf("%d%d%d", &c, &x, &y);
		if(c == 0) update(x, y);
		else printf("%d\n", query(x, y + 1, 1, 0, size_));
	}
}