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 (judge verdict: Time Limit Exceeded)

#include <stdio.h>

#define NMAX 100000
#define INIT 2147483647
int n;
int array[NMAX];

void init_array(void);
void update(int x, int y);
void find(int x, int y);

int main(){
   int iq, nq, cmd, x, y;
   scanf("%d %d",&n,&nq);
   init_array();
   for(iq=0;iq<nq;iq++){
      scanf("%d %d %d",&cmd,&x,&y);
      switch(cmd){
      case 0 :
         update(x,y);
         break;
      case 1 :
         find(x,y);
         break;
      }
   }
   return 0;
}

void init_array(void){
   int i;
   for(i=0;i<n;i++){
      array[i]=INIT;
   }
}

void update(int x, int y){
   array[x]=y;
}

void find(int x, int y){
   int i, min;
   min=array[x];
   for(i=x+1;i<=y;i++){
      if(array[i]<min) min=array[i];
   }
   printf("%d\n",min);
}