import java.util.*;
 
class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
 
        // Read the size of the array
        int n = sc.nextInt();
        int[] arr = new int[n];
 
        // Read the elements of the array
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
        }
 
        // Read the target sum k
        int k = sc.nextInt();
 
        // Initialize variables
        int currentSum = 0;
        int smallestLength = Integer.MAX_VALUE;
        int count = 0;
 
        // Map to store the frequency of prefix sums
        Map<Integer, Integer> map = new HashMap<>();
 
        // Initialize the map with the prefix sum of 0 coz what if i get k=5, and arr[0]=5
        map.put(0, -1);
 
        for (int i = 0; i < n; i++) {
            // Update the current prefix sum
            currentSum += arr[i];
 
            // Check if there exists a prefix sum that gives us the required sum k
            if (map.containsKey(currentSum - k)) {
                int length = i - map.get(currentSum - k);
 
                // Update the smallest length and count
                if (length < smallestLength) {
                    smallestLength = length;
                    count = 1; // Reset count for new smallest length
                } else if (length == smallestLength) {
                    count++; // Increment count for same smallest length
                }
            }
 
            // Update the frequency of the current prefix sum
            map.put(currentSum, i); // Store the latest index of the prefix sum
            /*
            say 1 2 3 4 5
            at end 15 so now i want length (i+1(0 indexed)) - map.containskey(currsum, i+1)
            jth index - i+1 where subarray sum starts but the i+1 in the map is solely coz of 0 indexing
            but i didnt like how it made lengths of 1 to become 0 so i made it i itself and put the map 
            start at (0,-1) prefix sum of 0 is at -1 so it doesnt exist yet
            */
        }
 
        // Output the count of special subarrays
        System.out.println(count == 0 ? 0 : count);
    }
}
 

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NQo1IDEgMiAzIDQKNQ==

5
5 1 2 3 4
5