Kolya is going to make fresh orange juice. He has n oranges of sizes a1, a2, ..., an. Kolya will put them in the juicer in the fixed order, starting with orange of size a1, then orange of size a2 and so on. To be put in the juicer the orange must have size not exceeding b, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.

The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than d. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?

Input

The first line of the input contains three integers n, b and d (1 ≤ n ≤ 100 000, 1 ≤ b ≤ d ≤ 1 000 000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value d, which determines the condition when the waste section should be emptied.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1 000 000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.

Output

Print one integer — the number of times Kolya will have to empty the waste section.

Examples

Input

2 7 10
5 6


Output

1


Input

1 5 10
7


Output

0


Input

3 10 10
5 7 7


Output

1


Input

1 1 1
1


Output

0

Note

In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.

In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all.
import java.util.*;
import java.io.*;

public class Solution{
	public Solution(){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int b = sc.nextInt();
		int d = sc.nextInt();
		
		int total = 0;
		int orange = 0;
		int res = 0;
		
		for(int i=0; i<n; i++){
			orange = sc.nextInt();
			
			if(orange <= b){
				total += orange;
			}
			
			if(total > d){
				total = 0;
				res++;
			}
		}
		
		System.out.println(res);
	}
	
	int binarySearch(int arr[], int l, int r, int x) 
    { 
        if (r >= l) { 
            int mid = l + (r - l) / 2; 
  
            // If the element is present at the 
            // middle itself 
            if (arr[mid] == x) 
                return mid; 
  
            // If element is smaller than mid, then 
            // it can only be present in left subarray 
            if (arr[mid] > x) 
                return binarySearch(arr, l, mid - 1, x); 
  
            // Else the element can only be present 
            // in right subarray 
            return binarySearch(arr, mid + 1, r, x); 
        } 
  
        // We reach here when element is not present 
        // in array 
        return -1; 
    }
	
	int partition(int arr[], int[] dep, int low, int high) 
    { 
        int pivot = arr[high];  
        int i = (low-1); // index of smaller element 
        for (int j=low; j<high; j++) 
        { 
            // If current element is smaller than or 
            // equal to pivot 
            if (arr[j] <= pivot) 
            { 
                i++; 
  
                // swap arr[i] and arr[j] 
                int temp = arr[i]; 
                arr[i] = arr[j]; 
                arr[j] = temp;
				
				if(dep != null){
					temp = dep[i];
					dep[i] = dep[j];
					dep[j] = temp;
				}
            } 
        } 
  
        // swap arr[i+1] and arr[high] (or pivot) 
        int temp = arr[i+1]; 
        arr[i+1] = arr[high]; 
        arr[high] = temp;
		
		if(dep != null){
			temp = dep[i+1];
			dep[i+1] = dep[high];
			dep[high] = temp;
		}
  
        return i+1; 
    } 
  
  
    /* The main function that implements QuickSort() 
      arr[] --> Array to be sorted, 
      low  --> Starting index, 
      high  --> Ending index */
    void sort(int arr[], int[] dep, int low, int high) 
    { 
        if (low < high) 
        { 
            /* pi is partitioning index, arr[pi] is  
              now at right place */
            int pi = partition(arr, dep, low, high); 
  
            // Recursively sort elements before 
            // partition and after partition 
            sort(arr, dep, low, pi-1); 
            sort(arr, dep, pi+1, high); 
        } 
    }
	
	public static void main(String[] args){
		new Solution();
	}
}