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 Python (3.4.3) (judge verdict: Runtime Error)

H, W = map(int, input().split())
s = [0 for _ in range(H)]
for i in range(H):
    s[i] = input()      # intをかけずにinputすると文字列として扱われる
# ここまでが入力

dx = [1, 0, 0, -1]    # これらは移動を表すリスト
dy = [0, 1, -1, 0]    # 順番に、下・右・左・上となっている

ans = "Yes"
for i in range(H):
    for j in range(W):
        if s[i][j] == ".":      # "#"が隣接しているかどうかが知りたいので、"."の周りに何があってもどうでもいい
            continue            # つまり注目しているマスが"#"のときだけ以下の処理をする
        ans = "No"      # この文を入れる場所は要注意
        for d in range(4):      # s[i][j]に対して、隣接する8マスについて"#"があるかを判定する
            di = i + dx[d]
            dj = j + dy[d]
            if di<0 or dj<0 or H-1<di or W-1<dj:    # 注目した隣接マスが「ない」ときは処理をしない
                continue
            if s[di][dj] == "#":
                ans = "Yes"
print(ans)