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 sys;

def init(n_):
    length = 1;
    while(length < n_):
        length*=2
    element=[0]*(2*length-1)
    for i in range(2*length-1):
        element[i] = initial
    return element,length

def update(k,a,element):
    k+=n-1
    element[k]=a
    while(k > 0):
        k = (int)((k-1)/2)
        element[k] = min(element[k*2+1],element[k*2+2])

def findmin(a, b):
    return findquery(a,b+1,0,0,n)

def findquery(a, b, k, l, r):
    if(r <= a or b <= l):
        return initial
    if(a <= l and r <= b):
        return element[k]

    vl = findquery(a,b,k*2+1,l,(int)((l+r)/2+0.5))
    vr = findquery(a,b,k*2+2,(int)((l+r)/2),r)
    return min(vl,vr)

n,q = map(int, input().split());

initial = pow(2,31)-1
element,n = init(n)

for i in range(q):
    query,key,value = map(int, input().split());
    if(query == 0):
        update(key,value,element)
    else :
        print(findmin(key,value))