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.*; 
import java.util.*; 
//import javafx.util.*; 
import java.math.*;
//import java.lang.*;
 
public class untitled 
{ 
    static class Pair{
        int x;int y;int i;
        Pair(int x,int y,int i){
            this.x=x;
            this.y=y;
            this.i=i;
        }
    }
    static class sorting implements Comparator<Pair>{
        public int compare(Pair a,Pair b){
            //return (a.y)-(b.y);
            return (a.x-b.x);
        }
    }
 
     static class Reader 
    { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  
        public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public Reader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public String readLine() throws IOException 
        { 
            byte[] buf = new byte[64]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 
  
        public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9'); 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
  
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
  
            if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            } 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        } 
  
        private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        } 
  
        public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
    } 
    
 static void merge(int arr[], int l, int m, int r) 
    { 
        // Find sizes of two subarrays to be merged 
        int n1 = m - l + 1; 
        int n2 = r - m; 
  
        /* Create temp arrays */
        int L[] = new int [n1]; 
        int R[] = new int [n2]; 
  
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i) 
            L[i] = arr[l + i]; 
        for (int j=0; j<n2; ++j) 
            R[j] = arr[m + 1+ j]; 
  
  
        /* Merge the temp arrays */
  
        // Initial indexes of first and second subarrays 
        int i = 0, j = 0; 
  
        // Initial index of merged subarry array 
        int k = l; 
        while (i < n1 && j < n2) 
        { 
            if (L[i] <= R[j]) 
            { 
                arr[k] = L[i]; 
                i++; 
            } 
            else
            { 
                arr[k] = R[j]; 
                j++; 
            } 
            k++; 
        } 
  
        /* Copy remaining elements of L[] if any */
        while (i < n1) 
        { 
            arr[k] = L[i]; 
            i++; 
            k++; 
        } 
  
        /* Copy remaining elements of R[] if any */
        while (j < n2) 
        { 
            arr[k] = R[j]; 
            j++; 
            k++; 
        } 
    } 
  
    // Main function that sorts arr[l..r] using 
    // merge() 
    static void sort(int arr[], int l, int r) 
    { 
        if (l < r) 
        { 
            // Find the middle point 
            int m = (l+r)/2; 
  
            // Sort first and second halves 
            sort(arr, l, m); 
            sort(arr , m+1, r); 
  
            // Merge the sorted halves 
            merge(arr, l, m, r); 
        } 
    }
   // static int n;static ArrayList<Pair> arr1;
    //static int len;
    public static int[] na(int n)throws IOException{
        Reader s1=new Reader();
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = s1.nextInt();
        return a;
    }
    public static void main(String[] args) throws IOException { 
       // Reader s=new Reader();
        Scanner s=new Scanner(System.in);
        int n=s.nextInt();
        s.nextLine();
        Stack<Integer> q=new Stack<>();
        long ans=0;
        long p=1;
        int check=1;
        long e=4294967295L;
        int count=0;
        for(int i=0;i<n;i++){
            String str=s.nextLine();
            if(str.substring(0,3).equals("add")){
                if(check==0){
                    System.out.println("OVERFLOW!!!");
                    return;
                }
                ans+=p;
                if(ans>e){
                    System.out.println("OVERFLOW!!!");
                    return;
                }
//                System.out.println(ans);
            }
            else if(str.substring(0,3).equals("end")){
                if(count!=0){
                    count--;
                    if(count==0){
                        check=1;
                    }
             //       System.out.println(count);
                }
                else if(count<1){
                    int a=q.pop();
               //     System.out.println(a);
                    p=p/a;
                    check=1;
                }
                

            }
            else{
            //    System.out.println(str.substring(4));
                if(check!=0){

                     p=p*Integer.parseInt(str.substring(4));
               //      System.out.println(p);
                     if(p>e){
                        check=0;

                    }
                     q.push(Integer.parseInt(str.substring(4)));
                }
                else{
                    count++;
      //              System.out.println(count+" 6666 ");
                }
            }
            
        }
        
       // System.out.println(a);
        if(ans>e){
            System.out.println("OVERFLOW!!!");

        }
        else{
        System.out.println(ans);
}
    }
// Returns factorial of n 
static int fact(int n) 
{ 
    int res = 1; 
    for (int i = 2; i <= n; i++) 
        res = res * i; 
    return res; 
}
    static boolean isPrime(int n) { 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
        if (n % 2 == 0 ||  
            n % 3 == 0) 
            return false; 
      
        for (int i = 5; 
                 i * i <= n; i = i + 6) 
            if (n % i == 0 || 
                n % (i + 2) == 0) 
                return false; 
      
        return true; 
    } 
    static int gcd(int a, int b) 
    { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }  
    static long fast_pow(long base,long n,long M)
    {
        if(n==0)
           return 1;
        if(n==1)
        return base;
        long halfn=fast_pow(base,n/2,M);
        if(n%2==0)
            return ( halfn * halfn ) % M;
        else
            return ( ( ( halfn * halfn ) % M ) * base ) % M;
    }
    static long findMMI_fermat(long n,int M)
    {
        return fast_pow(n,M-2,M);
    }
    
}