Educational Codeforces Round 26 A C 之Python

You are given a text of single-space separated words, consisting of small and capital Latin letters.

Volume of the word is number of capital letters in the word. Volume of the text is maximum volume of all words in the text.

Calculate the volume of the given text.

Input

The first line contains one integer number n (1 ≤ n ≤ 200) — length of the text.

The second line contains text of single-space separated words s₁, s₂, ..., s_i, consisting only of small and capital Latin letters.

Output

Print one integer number — volume of text.

Examples

Input

7
NonZERO

Output

Input

24
this is zero answer text

Output

Input

24
Harbour Space University

Output

Note

In the first example there is only one word, there are 5 capital letters in it.

In the second example all of the words contain 0 capital letters.

判断每个字符串中大写字母出现的次数的最大值

#python3
n, a = input(), input().split()
k = 0
for i in a:
    ans = 0
    for c in i:
        if c.isupper():
            ans += 1
    k = max(k, ans)
print(k)

One very important person has a piece of paper in the form of a rectangle a × b.

Also, he has n seals. Each seal leaves an impression on the paper in the form of a rectangle of the size x_i × y_i. Each impression must be parallel to the sides of the piece of paper (but seal can be rotated by 90 degrees).

A very important person wants to choose two different seals and put them two impressions. Each of the selected seals puts exactly one impression. Impressions should not overlap (but they can touch sides), and the total area occupied by them should be the largest possible. What is the largest area that can be occupied by two seals?

Input

The first line contains three integer numbers n, a and b (1 ≤ n, a, b ≤ 100).

Each of the next n lines contain two numbers x_i, y_i (1 ≤ x_i, y_i ≤ 100).

Output

Print the largest total area that can be occupied by two seals. If you can not select two seals, print 0.

Examples

Input

2 2 2
1 2
2 1

Output

Input

Output

Input

Output

Note

In the first example you can rotate the second seal by 90 degrees. Then put impression of it right under the impression of the first seal. This will occupy all the piece of paper.

In the second example you can't choose the last seal because it doesn't fit. By choosing the first and the third seals you occupy the largest area.

In the third example there is no such pair of seals that they both can fit on a piece of paper.

在n×m的矩形中可以放如两个矩形求使得面积最大的是多少

#python3
k, n, m = map(int,input().split())
x, y = [],[]
for i in range(k):
    q, p = map(int, input().split())
    x.append(min(q, p))
    y.append(max(q, p))
MAX = 0
for i in range(k):
    for j in range(i+1,k):
        ok = False
        if x[i] + x[j] <= n and y[i] <= m and y[j] <= m:
            ok = True
        elif x[i] + y[j] <= n and y[i] <= m and x[j] <= m:
            ok = True
        elif y[i] + y[j] <= m and x[i] <= n and x[j] <= n:
            ok = True
        elif y[i] + x[j] <= m and x[i] <= n and y[j] <= n:
            ok = True
        elif y[i] + y[j] <= n and x[i] <= m and x[j] <= m:
            ok = True
        elif y[i] + x[j] <= n and x[i] <= m and y[j] <= m:
            ok = True
        elif x[i] + x[j] <= m and y[i] <= n and y[j] <= n:
            ok = True
        elif x[i] + y[j] <= m and y[i] <= n and x[j] <= n:
            ok = True
        if ok:
            MAX = max(MAX, x[i]*y[i] + x[j]*y[j])
print(MAX)