Problem: You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:

  * for n — for loop; 
  * end — every command between "for n" and corresponding "end" is executed n times; 
  * add — adds 1 to x. 



After the execution of these commands, value of x is returned.

Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.

Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command. 

Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.

If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.

Input

The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.

Each of the next l lines contains a single command of one of three types:

  * for n (1 ≤ n ≤ 100) — for loop; 
  * end — every command between "for n" and corresponding "end" is executed n times; 
  * add — adds 1 to x. 

Output

If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.

Examples

Input


9
add
for 43
end
for 10
for 15
add
end
add
end


Output


161


Input


2
for 62
end


Output


0


Input


11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end


Output


OVERFLOW!!!

Note

In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.

In the second example there are no commands "add", thus the returning value is 0.

In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import java.util.*;
import java.io.PrintStream;
import java.io.PrintWriter;
 
public class A {
 
	static final int mod = 1000000007;
	static final int temp = 998244353;
	static final long MOD = 1000000007;
	static final long M = (long)1e9+7;
 
	
	static class Pair implements Comparable<Pair>
	{
		int first, second;
		public Pair(int aa, int bb)
		{
			first = aa; second = bb;
		}
		public int compareTo(Pair p)
		{
//			if(a == p.a) return b - p.b;
//			return a - p.a;
			if(first == p.first) return (int)(second - p.second);
			return (int)(first - p.first);
		}
	}
	
	/* IO for List of Arrays
	 * ArrayList<Integer>[] list = new ArrayList[m];
		for(int i = 0;i<m;i++)
		{
			list[i] = new ArrayList<>();
		}
	 */
	
	/*
	 * IO FOR 2D GRID IN JAVA
	 * 
	 * 		char[][] grid = new char[n][n];
	 * 		for(int i = 0;i<n;i++) grid[i] = sc.next().toCharArray();

	 * */
	
	static class Reader {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		List<Integer> readList(int n)
		{
			List<Integer> l = new ArrayList<>();
			for(int i = 0;i<n;i++) l.add(nextInt());
			return l;
		}
		
		long[] longReadArray(int n) throws IOException {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
 
	public static boolean isPrime(long n) {
		if(n == 1)
		{
			return false;
		}
		for(long i = 2;i*i<=n;i++)
		{
			if(n%i == 0)
			{
				return false;
			}
		}
		return true;
	}
	
	public static List<Integer> Sieve(int n)
	{
		boolean prime[] = new boolean[n+1];
		Arrays.fill(prime, true);
		List<Integer> l = new ArrayList<>();
		
		
		for (int p=2; p*p<=n; p++) 
		{ 
			if (prime[p] == true) 
			{ 	
			    for(int i=p*p; i<=n; i += p) 
				{
				    prime[i] = false; 
				}	
			} 
		} 
 
		for (int p=2; p<=n; p++) 
		{
		    if (prime[p] == true)
		    {
		       l.add(p); 
		    }
		}
		
		return l;
	}
	
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	} 
 
	public static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}
	
	public static long LongLCM(long a, long b)
    {
        return (a / LongGCD(a, b)) * b;
    }
	
	public static int phi(int n)  //euler totient function
    { 
        int result = 1; 
        for (int i = 2; i < n; i++) 
            if (gcd(i, n) == 1) 
                result++; 
        return result; 
    }
	
	    
	public static int[] computePrefix(int arr[], int n)
	{
		int[] prefix = new int[n];
		prefix[0] = arr[0];
		for(int i = 1;i<n;i++)
		{
			prefix[i] = prefix[i-1]+arr[i];
		}
		
		return prefix;
	}
	
	
	public static long fastPow(long x, long n) //include mod at each step if asked and in args of fn too
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	
	
	   public static long modPow(long x, long y, long p)
	    {
	 
	        long res = 1;
	 
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	 
	        return res;
	    }
	   
	   static long modInverse(long n, long p)
	    {
	        return modPow(n, p - 2, p);
	    }
	 
	    // Returns nCr % p using Fermat's little theorem.
	    
	    public static long nCr(long n, long r,long p)
	    {
	          if (n<r) 
	              return 0;
	          
	        if (r == 0)
	            return 1;
	 
	        long[] fac = new long[(int)(n) + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
	                % p * modInverse(fac[(int)(n - r)], p)
	                % p)
	            % p;
	    }
	    
	    public static long[] modInvArr(int n, long mod)
	    {
	    	long[] ans = new long[n+1];
			ans[1] = 1;
			for(int i = 2;i <= n;i++)
			{
				ans[i] = (mod - (mod/i) * ans[(int)(mod%i)])%mod;
			}
			return ans;
	    }
 
		public static void Sort(int[] a) {
			List<Integer> l=new ArrayList<>();
			for (int i:a) l.add(i);
			Collections.sort(l);
			//Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		public static void ssort(char[] a) {
			List<Character> l=new ArrayList<>();
			for (char ch : a) l.add(ch);
			Collections.sort(l);
			//Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		//Modular Operations for Addition and Multiplication. 
		
		   public static long perfomMod(long x){
		        return ((x%M + M)%M);
		    }
		   public static long addMod(long a, long  b){
		        return perfomMod(perfomMod(a)+perfomMod(b));
		    }
		   public static long mulMod(long  a, long b){
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }

		   
	public static void main(String[] args) throws IOException 
	{
		Reader sc=new Reader();
		PrintWriter out = new PrintWriter(System.out);

		int t = sc.nextInt();
		Stack<Long> stk = new Stack<>();
		stk.push(1L);
		long res = 0, INF = 1L<<32;
		while(t-- > 0)
		{
			String s = sc.next();
			if(s.equals("for"))
			{
				int x = sc.nextInt();
				long adder = min(INF,stk.peek() * x);
				stk.push(adder);
			}
			else if(s.equals("end"))
			{
				stk.pop();
			}
			else
			{
				res += stk.peek();
			}
		}
		
		if(res >= INF)
		{
			System.out.println("OVERFLOW!!!");
		}
		else
		{
			System.out.println(res);
		}
		
		out.close();
	}
 
	
}