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: Accepted)

import bisect
import collections
import heapq
import itertools
import operator
import sys


def main():
    int_max = 2 ** 31 - 1
    n, q = map(int, sys.stdin.readline().split())
    n = 1 << (n - 1).bit_length()
    dat = [int_max] * 131072 * 2

    def update(i, x):
        i += n - 1
        dat[i] = x
        while i > 0:
            i = (i - 1) // 2
            dat[i] = min(dat[i * 2 + 1], dat[i * 2 + 2])

    def query(a, b, k, l, r):
        if r <= a or b <= l:
            return int_max
        if a <= l and r <= b:
            return dat[k]
        else:
            vl = query(a, b, k * 2 + 1, l, (l + r) // 2)
            vr = query(a, b, k * 2 + 2, (l + r) // 2, r)
            return min(vl, vr)

    for _ in range(q):
        com, x, y = map(int, sys.stdin.readline().split())
        if com == 0:
            update(x, y)
        elif com == 1:
            print(query(x, y + 1, 0, 0, n))


if __name__ == "__main__":
    main()