#include <cstdio>
#include <algorithm>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <string>
#include <string.h>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#include <iostream>
#include <sstream>
#include <cctype>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < n; ++i)
#define rrep(i,n) for(int i = 1; i <= n; ++i)
#define drep(i,n) for(int i = n-1; i >= 0; --i)
#define gep(i,g,j) for(int i = g.head[j]; i != -1; i = g.e[i].next)
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) x = max(x,y)
#define mins(x,y) x = min(x,y)
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcount
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
typedef vector<int> vi;
inline int in() { int x; scanf("%d",&x); return x;}
inline void priv(vi& a) { rep(i,sz(a)) printf("%d%c",a[i],i==sz(a)-1?'\n':' ');}

const int MX = 100005, INF = 1000010000;
const ll LINF = 1000000000000000000ll;
const double eps = 1e-10;
const int di[] = {-1,0,1,0}, dj[] = {0,-1,0,1}; //^<v>

string s;int n;

struct Pa {
  string s; 
  bool f;
  int i;
  Pa(string s):s(s){}
  ll parse() {
    i = 0; f = true;
    ll x = exec();
    if (!f) return -LINF;
    return x;
  }
  ll exec() {
    ll x = num();
    if (!f) return 0;
    while (i < n) {
      if (s[i] == '+') {
        ++i;
        ll y = num();
        if (!f) return 0;
        x += y;
      } else if(s[i] == '-') {
        ++i;
        ll y = num();
        if (!f) return 0;
        x -= y;
      } else { f = false; return 0;}
    }
    return x;
  }
  ll num() {
    if (!isdigit(s[i])) {
      f = false;
      return 0;
    }
    ll x = s[i++]-'0';
    while (i < n && isdigit(s[i])) {
      x = x*10+s[i++]-'0';
    }
    return x;
  }
};

int main(){
  cin >> s;
  n = sz(s);
  s += s;
  ll ans = -LINF;
  rep(i,n) {
    Pa p(s.substr(i,n));
    ans = max(ans,p.parse());
  }
  cout << ans << endl;
  return 0;
}