You have a string of decimal digits s. Let's define bij = si·sj. Find in matrix b the number of such rectangles that the sum bij for all cells (i, j) that are the elements of the rectangle equals a in each rectangle.

A rectangle in a matrix is a group of four integers (x, y, z, t) (x ≤ y, z ≤ t). The elements of the rectangle are all cells (i, j) such that x ≤ i ≤ y, z ≤ j ≤ t.

Input

The first line contains integer a (0 ≤ a ≤ 109), the second line contains a string of decimal integers s (1 ≤ |s| ≤ 4000).

Output

Print a single integer — the answer to a problem.

Please, do not write 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

10
12345


Output

6


Input

16
439873893693495623498263984765


Output

40
from collections import Counter
a=int(input())
string=input()
arr=[0]
count1=0
count2=0
for i in range(len(string)):
  arr+=[int(string[i])]
for i in range(1,len(arr)):
  arr[i]+=arr[i-1]
#print(arr)
temparr=[]
#sumset.add(arr[0])
for i in range(len(arr)):
  for j in range(i+1,len(arr)):
    temparr+=[(arr[j]-arr[i])]
sumset=Counter(temparr)
#print(sumset)
possums=0
for i in (sumset):
  possums+=sumset[i]
for i in (sumset):
  #print(str(i)+"###")
  if i!=0 and a%i==0 and i**2!=a:
    count1+=sumset[i]*sumset[a//i]
  elif i==0 and a==0:
    count1+=sumset[i]*possums
  elif i!=0 and a%i==0 and i**2==a:
    #print(str(i)+"***")
    count2+=sumset[i]*sumset[a//i]

print(count1+count2)