Given a matrix (H × W) which contains only 1 and 0, find the area of the largest rectangle which only contains 0s.

Constraints

* 1 ≤ H, W ≤ 1,400

Input


H W
c1,1 c1,2 ... c1,W
c2,1 c2,2 ... c2,W
:
cH,1 cH,2 ... cH,W


In the first line, two integers H and W separated by a space character are given. In the following H lines, ci,j, elements of the H × W matrix, are given.

Output

Print the area (the number of 0s) of the largest rectangle.

Example

Input

4 5
0 0 1 0 0
1 0 0 0 0
0 0 0 1 0
0 0 0 1 0


Output

6

"""
Writer: SPD_9X2
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DPL_3_B&lang=ja

各セルに対して、上にいくつ0が連続しているか数えれば、
ヒストグラム上での最大長方形問題になる
"""


def Largest_rectangle_in_histgram(lis):

    stk = []

    ans = 0
    N = len(lis)

    for i in range(N):

        if len(stk) == 0:
            stk.append((lis[i],i))

        elif stk[-1][0] < lis[i]:
            stk.append((lis[i],i))

        elif stk[-1][0] == lis[i]:
            pass

        else:
            lastpos = None
            while len(stk) > 0 and stk[-1][0] > lis[i]:

                nh,np = stk[-1]
                lastpos = np
                del stk[-1]

                ans = max(ans , nh*(i-np))

            stk.append((lis[i] , lastpos))

    return ans

H,W = map(int,input().split())

c = []
for i in range(H):
    C = list(map(int,input().split()))
    c.append(C)

zlis = [ [0] * (W+1) for i in range(H) ]
for i in range(H):
    for j in range(W):

        if c[i][j] == 1:
            continue
        elif i == 0:
            zlis[i][j] = 1
        else:
            zlis[i][j] = zlis[i-1][j] + 1

#print (zlis)
ans = 0
for i in range(H):
    ans = max(ans , Largest_rectangle_in_histgram(zlis[i]))

print (ans)

