A string is binary, if it consists only of characters "0" and "1".

String v is a substring of string w if it has a non-zero length and can be read starting from some position in string w. For example, string "010" has six substrings: "0", "1", "0", "01", "10", "010". Two substrings are considered different if their positions of occurrence are different. So, if some string occurs multiple times, we should consider it the number of times it occurs.

You are given a binary string s. Your task is to find the number of its substrings, containing exactly k characters "1".

Input

The first line contains the single integer k (0 ≤ k ≤ 106). The second line contains a non-empty binary string s. The length of s does not exceed 106 characters.

Output

Print the single number — the number of substrings of the given string, containing exactly k characters "1".

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

1
1010


Output

6


Input

2
01010


Output

4


Input

100
01010


Output

0

Note

In the first sample the sought substrings are: "1", "1", "10", "01", "10", "010".

In the second sample the sought substrings are: "101", "0101", "1010", "01010".
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
	public static long mod= (long) (1e9 +7);
	public static void main(String args[])
	{
		InputReader s= new InputReader(System.in);
		OutputStream outputStream= System.out;
		PrintWriter out= new PrintWriter(outputStream);
		int k= s.nextInt();
		String str= s.next();
		int n= str.length();
		int sum[]= new int[n];
		sum[0] = str.charAt(0)-'0';
		for(int i=1;i<n;i++){
			sum[i] = sum[i-1] + str.charAt(i)-'0';
		}
		int dp[]= new int[n+1];
		dp[0]=1;
		long ans=0;
		for(int i=0;i<n;i++){
			if(sum[i]>=k){
				ans+= dp[sum[i]-k];
			}
			dp[sum[i]]++;
		}
		out.println(ans);
		out.close();
	}
	
	static long combinations(long n,long r){   // O(r)
		if(r==0 || r==n) return 1;
		r= Math.min(r, n-r);
		long ans=n;  // nC1=n;
		for(int i=1;i<r;i++){
			ans= ans*(n-i);
			ans= ans/(i+1);
		}
		return ans;
		
	}
	public static void debug(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}
	static long modulo(long a,long b,long c) {
		long x=1;
		long y=a;
		while(b > 0){
			if(b%2 == 1){
				x=(x*y)%c;
			}
			y = (y*y)%c; // squaring the base
			b /= 2;
		}
		return  x%c;
	}
	static void catalan_numbers(int n) {
		long catalan[]= new long[n+1];
	    catalan[1] = 1;
	    for (int i = 2; i <= n; i++) {
	        for (int j = 1; j <= i - 1; j++) {
	            catalan[i] = catalan[i] + ((catalan[j]) * catalan[i - j]);
	        }
	    }
	}
	
	static ArrayList<Integer> primeFactors(int n)   // O(sqrt(n))
	{
	    // Print the number of 2s that divide n
		ArrayList<Integer> arr= new ArrayList<>();
	    while (n%2 == 0)
	    {
	        arr.add(2);
	        n = n/2;
	    }
	 
	    // n must be odd at this point.  So we can skip one element (Note i = i +2)
	    for (int i = 3; i <= Math.sqrt(n); i = i+2)
	    {
	        // While i divides n, print i and divide n
	        while (n%i == 0)
	        {
	            arr.add(i);
	            n = n/i;
	        }
	    }
	    // This condition is to handle the case when n is a prime number
	    // greater than 2
	    if (n > 2)
	        arr.add(n);
	    
	    return arr;
	}

	public static long expo(long a, long b){
		if (b==0)
	        return 1%mod;
	    if (b==1)
	        return a%mod;
	    if (b==2)
	        return ((a%mod)*(a%mod))%mod;

	    if (b%2==0){
	            return expo(expo(a%mod,(b%mod)/2)%mod,2%mod)%mod;
	    }
	    else{
	        return (a%mod)*(expo(expo(a%mod,(b-1)%mod/2)%mod,2%mod)%mod)%mod;
	    }
	}
	static class Pair implements Comparable<Pair>
	{
		long f;
		String s;
		Pair(long ii, String cc)
		{
			f=ii;
			s=cc;
		}
		
		public int compareTo(Pair o) 
		{
			return Long.compare(this.f, o.f);
		}
		
	}
	
	public static int[] sieve(int N){    // O(n*log(logn))
		int arr[]= new int[N+1];
		for(int i=2;i<=Math.sqrt(N);i++){
			if(arr[i]==0){
				for(int j= i*i;j<= N;j= j+i){
					arr[j]=1;
				}
			}
		}
		return arr;
		// All the i for which arr[i]==0 are prime numbers.
	}
	static long gcd(long a,long b){       // O(logn)
		if(b==0) return a;
		return gcd(b,a%b);
	}
	static class InputReader {

	    public BufferedReader reader;
	    public StringTokenizer tokenizer;
 
	    public InputReader(InputStream inputstream) {
	      reader = new BufferedReader(new InputStreamReader(inputstream));
	      tokenizer = null;
	    }
	    
	    public String nextLine(){
	    	String fullLine=null;
	    	while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	              fullLine=reader.readLine();
	            } catch (IOException e) {
	              throw new RuntimeException(e);
	            }
	            return fullLine;
	          }
	          return fullLine;
	    }
		public String next() {
	      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	        try {
	          tokenizer = new StringTokenizer(reader.readLine());
	        } catch (IOException e) {
	          throw new RuntimeException(e);
	        }
	      }
	      return tokenizer.nextToken();
	    }
		public long nextLong() {
		      return Long.parseLong(next());
		    }
	    public int nextInt() {
	      return Integer.parseInt(next());
	    }
	  }
}