Zookeeper is buying a carton of fruit to feed his pet wabbit. The fruits are a sequence of apples and oranges, which is represented by a binary string s_1s_2… s_n of length n. 1 represents an apple and 0 represents an orange.

Since wabbit is allergic to eating oranges, Zookeeper would like to find the longest contiguous sequence of apples. Let f(l,r) be the longest contiguous sequence of apples in the substring s_{l}s_{l+1}… s_{r}. 

Help Zookeeper find ∑_{l=1}^{n} ∑_{r=l}^{n} f(l,r), or the sum of f across all substrings.

Input

The first line contains a single integer n (1 ≤ n ≤ 5 ⋅ 10^5).

The next line contains a binary string s of length n (s_i ∈ \{0,1\}) 

Output

Print a single integer: ∑_{l=1}^{n} ∑_{r=l}^{n} f(l,r). 

Examples

Input


4
0110


Output


12


Input


7
1101001


Output


30


Input


12
011100011100


Output


156

Note

In the first test, there are ten substrings. The list of them (we let [l,r] be the substring s_l s_{l+1} … s_r):

  * [1,1]: 0 
  * [1,2]: 01 
  * [1,3]: 011 
  * [1,4]: 0110 
  * [2,2]: 1 
  * [2,3]: 11 
  * [2,4]: 110 
  * [3,3]: 1 
  * [3,4]: 10 
  * [4,4]: 0 



The lengths of the longest contiguous sequence of ones in each of these ten substrings are 0,1,2,2,1,2,2,1,1,0 respectively. Hence, the answer is 0+1+2+2+1+2+2+1+1+0 = 12.

// Imports
import java.util.*;
import java.io.*;

public class F1428 {

    // Binary Indexed Tree based off of template used in past problems
    static class BIT {
        public long[] bit;
        
        public BIT(int n) {
            bit = new long[n + 3];
        }
        
        public void update(int index, long val) {
            index++;
        
            while(index < bit.length) {
                bit[index] += val;
                index += index & -index;
            }
        }
        
        public long query(int index) {
            // System.out.println("QUERY: index=" + index);
            long ret = 0;
            index++;
        
            while(index > 0) {
                ret += bit[index];
                index -= index & -index;
            }
            
            // System.out.println(ret);
            return ret;
        }
        
    }
    
    /**
     * @param args the command line arguments
     * @throws IOException, FileNotFoundException 
     */
    public static void main(String[] args) throws IOException, FileNotFoundException {
        
        // TODO UNCOMMENT WHEN ALGORITHM CORRECT
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        
        /* BufferedReader f = new BufferedReader(new StringReader("5\n" +
"11111")); */
        // [1, 1] "1"- 1
        // [1, 2] - 1
        // [1, 3] - 1
        // [2, 2] - 0
        // [2, 3] - 1
        // [3, 3] - 1
        // = 5
        
        // 10111
        // 0 / 1 -- 2
        
        // TODO code application logic here
        int N = Integer.parseInt(f.readLine());
        String s = f.readLine();
        BIT curr = new BIT(N);
        
        long sum = 0;
        long lastdiff = 0;
        int lastzero = -1;
        
        for(int i = 0; i < N; i++) {
            // System.out.println("i=" + i + " sum=" + sum);
            if(s.charAt(i) == '0') {
                sum += lastdiff;
                // Add in the value at position 0
                curr.update(0, 1);
                lastzero = i;
            }
            else {
                int diff = i - lastzero;
                // System.out.println("DIFF: " + diff);
                long onlyone = curr.query(diff - 1) - curr.query(diff - 2);
                lastdiff += curr.query(diff - 1);
                lastdiff++;
                
                // Add to sum all values under a certain value
                sum += lastdiff;
                curr.update(diff, onlyone);
                curr.update(diff - 1, -onlyone);
                curr.update(Math.max(1, diff-1), 1);
            }
        }
        
        System.out.println(sum);
    }

}
