class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        stack<int> s;
        const int n = heights.size();
        int *L = new int[n+1]{0};
        int *R = new int[n+1]{0};
        for(int i = 0;i < n;++i) {
            while(s.size() && heights[s.top()] > heights[i]) {
                R[s.top()] = i;
                s.pop();
            }
            if(s.empty()) L[i] = -1;
            else L[i] = s.top();
            s.push(i);
        }
        while(!s.empty()) {
            R[s.top()] = n;
            s.pop();
        }
        int ans = 0, tmp;
        for(int i = 0;i < n;++i) {
            tmp = heights[i]*(R[i]-L[i]-1);
            if(tmp > ans) ans = tmp;
        }
        return ans;
    }
};

#include <iostream>
#include <stack>
#include <algorithm>

#define npos (n+1)

using namespace std;
typedef long long longs;

const int N = 1e5+5;
int h[N],l[N],r[N];
longs ans;

inline void pop(stack<int> &s, int i)
{
    r[s.top()] = i;
    s.pop();
}

inline void push(stack<int> &s, int i, int a[])
{
    while(s.size()&&a[s.top()]>a[i])pop(s,i);
    if(s.empty()) l[i] = 0;
    else l[i] = s.top();
    s.push(i);
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    while(cin>>n&&n)
    {
        stack<int> s;
        ans = 0;
        for(int i=1;i<=n;++i)
        {
            cin>>h[i];
            push(s,i,h);
        }
        while(!s.empty())
            pop(s,npos);
        for(int i=1;i<=n;++i)
            ans = max(ans,(longs)h[i]*(r[i]-l[i]-1));
        cout<<ans<<endl;
    }

    return 0;
}

#include <iostream>
#include <stack>
#include <algorithm>

using namespace std;
typedef long long longs;

struct item{int w,h;};      // w是宽度，h是高度

const int N = 1050;
int n,m;
char in;
int tmph[N]{0};             // 可维持的最好高度
int ans = 0;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    stack<item> s;          // 严格递减的单调栈

    cin>>n>>m;
    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<=m;++j)
        {
            cin>>in;
            if(in=='F') ++tmph[j];
            else tmph[j] = 0;
            int tmpw = 0;   // 可维持的最好宽度
            while(!s.empty()&&s.top().h>=tmph[j])   // 可以延长
            {
                tmpw += s.top().w;                  // 外延宽度
                ans = max(ans,s.top().h*tmpw);
                s.pop();
            }
            s.push({tmpw+1,tmph[j]});        // 增加自己的宽度
        }
        
        int tmpw = 0;
        while(!s.empty())   // 出来时最好高度是严格递减的，均可外延
        {
            tmpw += s.top().w;
            ans = max(ans,s.top().h*tmpw);          // 反向外延更新答案
            s.pop();
        }
    }
    longs out = 3ll*ans;
    cout<<out<<endl;

    return 0;
}