You are given a matrix consisting of digits zero and one, its size is n × m. You are allowed to rearrange its rows. What is the maximum area of the submatrix that only consists of ones and can be obtained in the given problem by the described operations?

Let's assume that the rows of matrix a are numbered from 1 to n from top to bottom and the columns are numbered from 1 to m from left to right. A matrix cell on the intersection of the i-th row and the j-th column can be represented as (i, j). Formally, a submatrix of matrix a is a group of four integers d, u, l, r (1 ≤ d ≤ u ≤ n; 1 ≤ l ≤ r ≤ m). We will assume that the submatrix contains cells (i, j) (d ≤ i ≤ u; l ≤ j ≤ r). The area of the submatrix is the number of cells it contains.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 5000). Next n lines contain m characters each — matrix a. Matrix a only contains characters: "0" and "1". Note that the elements of the matrix follow without any spaces in the lines.

Output

Print a single integer — the area of the maximum obtained submatrix. If we cannot obtain a matrix of numbers one, print 0.

Examples

Input

1 1
1


Output

1


Input

2 2
10
11


Output

2


Input

4 3
100
011
000
101


Output

2
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007, INF = 2000000001;
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1, 0},
    dy[] = {-1, 0, 1, 0, -1, 1, 1, -1, 0};
const double EPS = 1e-9;
void init(string fname) {
  freopen((fname + ".in").c_str(), "r", stdin);
  freopen((fname + ".out").c_str(), "w", stdout);
}
long long gcd(long long a, long long b) { return (b != 0) ? gcd(b, a % b) : a; }
long long modpow(long long a, long long x, long long m = mod) {
  int p = 1;
  while (x > 0) {
    if (x & 1 == 1) {
      p = p * a % m;
      --x;
    } else {
      x >>= 1;
      a = a * a % m;
    }
  }
  return p;
}
void getZ(const string& s, vector<int>& v) {
  v = vector<int>(s.length());
  int n = s.length();
  for (int i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) v[i] = min(v[i - l], r - i + 1);
    while (i + v[i] < n && s[v[i]] == s[i + v[i]]) ++v[i];
    if (i + v[i] - 1 > r) r = (l = i) + v[i] - 1;
  }
}
long long dfs(vector<vector<int> >& g, vector<long long>& a, int v,
              vector<bool>& used, vector<long long>& pop) {
  used[v] = true;
  long long ans = a[v];
  for (int i = 0; i < g[v].size(); ++i) {
    if (!used[g[v][i]]) {
      ans += dfs(g, a, g[v][i], used, pop);
    }
  }
  return pop[v] = ans;
}
void make_set(vector<int>& p, vector<int>& r, int v) {
  p[v] = v;
  r[v] = 0;
}
int find_set(vector<int>& p, int v) {
  return p[v] == v ? v : p[v] = find_set(p, p[v]);
}
void unite_sets(vector<int>& p, vector<int>& r, int v1, int v2) {
  int p1 = v1, p2 = v2;
  if (r[p1] > r[p2]) swap(p1, p2);
  p[p1] = p2;
  if (r[p1] == r[p2]) ++r[p2];
}
class Edge {
 public:
  double weight;
  int v1, v2;
  Edge(double w = 0, int i = 0, int j = 0) : weight(w), v1(i), v2(j) {}
};
class CompEdge {
 public:
  bool operator()(const Edge& a, const Edge& b) const {
    return (a.weight < b.weight) ||
           (a.weight == b.weight &&
            (a.v1 < b.v1 || (a.v1 == b.v1 && a.v2 < b.v2)));
  }
};
long long gcdex(long long a, long long b, long long& x, long long& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  } else {
    long long x1, y1;
    long long g = gcdex(b % a, a, x1, y1);
    x = y1 - b / a * x1;
    y = x1;
    return g;
  }
}
void _a() {}
void _b() {}
void _c() {}
void _d() {
  int n, m;
  cin >> n >> m;
  vector<string> matrix(n);
  for (int i = 0; i < n; ++i) cin >> matrix[i];
  vector<vector<int> > r(n, vector<int>(m + 1));
  for (int i = n - 1; i >= 0; --i) {
    for (int j = m - 1; j >= 0; --j) {
      r[i][j] = (matrix[i][j] == '0') ? 0 : r[i][j + 1] + 1;
    }
  }
  int ans = 0;
  vector<int> t(n);
  for (int j = 0; j < m; ++j) {
    for (int i = 0; i < n; ++i) {
      t[i] = r[i][j];
    }
    sort(t.begin(), t.end());
    for (int i = 0; i < n; ++i) ans = max(ans, (n - i) * t[i]);
  }
  cout << ans;
}
void _e() {}
int main() {
  _d();
  return 0;
}
