Problem:
Score: 100 points

Problem Statement

In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year.

For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year.

In this country, a date is called Takahashi when the month and the day are equal as numbers. For example, 5-5 is Takahashi.

How many days from 2018-1-1 through 2018-a-b are Takahashi?

Constraints

a is an integer between 1 and 12 (inclusive).

b is an integer between 1 and 31 (inclusive).

2018-a-b is a valid date in Gregorian calendar.

Input

Input is given from Standard Input in the following format:

a b

Output

Print the number of days from 2018-1-1 through 2018-a-b that are Takahashi.

Sample Input 1

5 5

Sample Output 1

5

There are five days that are Takahashi: 1-1, 2-2, 3-3, 4-4 and 5-5.

Sample Input 2

2 1

Sample Output 2

1

There is only one day that is Takahashi: 1-1.

Sample Input 3

11 30

Sample Output 3

11

There are eleven days that are Takahashi: 1-1, 2-2, 3-3, 4-4, 5-5, 6-6, 7-7, 8-8, 9-9, 10-10 and 11-11.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){

  int H,W;
  int p,q;
  int i,j;
  int N=0;
  int Flag = 0;
  int rslt = 0;
  char result[3];
  
  scanf("%d %d",&H,&W);


  int F[H+2][W+2];
  char tmp[W];

  for(i=0;i<H+2;i++){
    for(j=0;j<W+2;j++){
      F[i][j] = '.';
    }
  }


  for(i=1;i<H+1;i++){
    scanf("%s",tmp);
    for(j=1;j<W+1;j++){
      F[i][j] = tmp[j];
    }
  }

  for(i=1;i<H+1;i++){
    for(j=1;j<W+1;j++){
      if(F[i][j] == '#'){
        if(F[i][j-1] == '.') Flag++;
        if(F[i][j+1] == '.') Flag++;
        if(F[i+1][j] == '.') Flag++;
        if(F[i-1][j] == '.') Flag++;
        if(Flag == 4)rslt = 1;
        Flag = 0;
        N++;
      }
    }
  }


  

  if(rslt == 1){
    printf("No");
  }
  else{
    printf("Yes");
  }
  

  return 0;
}