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 Python3 (judge verdict: Time Limit Exceeded)

def update(a, i, x) :
    a[i] = x


def find(a, s, t) :
    return min(a[s:t+1])


def main() :
    n, q = [int(i) for i in input().split()]
    a = [2**31 - 1]*n
    com = []
    x = []
    y= []
    for _ in range(q) :
        tmp = list(map(int, input().split()))
        com.append(tmp[0])
        x.append(tmp[1])
        y.append(tmp[2])

    for i in range(q) :
        if com[i] == 0 :
            update(a, x[i], y[i])
        elif com[i] == 1 :
            print(find(a, x[i], y[i]))


if __name__ == '__main__' :
    main()