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 Ruby (judge verdict: Wrong Answer)

s=STDIN.gets.split
n = s[0].to_i
q = s[1].to_i
int_max=2**31-1
ar  = Array.new(100001,(int_max))
ar2 = Array.new(1001  ,(int_max))

1.upto(q) {
s=STDIN.gets.split
com = s[0].to_i
x   = s[1].to_i
y   = s[2].to_i
#printf(":%d %d %d\n",com,x,y)
if(com==0)
  ar[x]=y
  ar2[x/100]=y if ar2[x/100] > y
elsif (com==1)
  min_=int_max
  x1=x/100
  y1=y/100
  #printf("x1y1=%d %d\n",x1,y1)
  if(x1==y1)
    x.upto(y){|i|
      min_ = ar[i] if min_ > ar[i]
    }
  else
    x1+=1 if (x % 100)>0
    x.upto(x1*100-1){|i|
      min_ = ar[i] if min_ > ar[i]
    }
    x1.upto(y1){|i|
      min_ = ar2[i] if min_ > ar2[i]
    }
    (y1*100).upto(y){|i|
      min_ = ar[i] if min_ > ar[i]
    }
  end
  puts min_
end
}