The bear has a string *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">|*s*|</sub> (record |*s*| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices *i*, *j* (1 ≤ *i* ≤ *j* ≤ |*s*|), that string *x*(*i*, *j*) = *s*<sub class="lower-index">*i*</sub>*s*<sub class="lower-index">*i* + 1</sub>... *s*<sub class="lower-index">*j*</sub> contains at least one string "bear" as a substring.

String *x*(*i*, *j*) contains string "bear", if there is such index *k* (*i* ≤ *k* ≤ *j* - 3), that *s*<sub class="lower-index">*k*</sub> = *b*, *s*<sub class="lower-index">*k* + 1</sub> = *e*, *s*<sub class="lower-index">*k* + 2</sub> = *a*, *s*<sub class="lower-index">*k* + 3</sub> = *r*.

Help the bear cope with the given problem.
The first line contains a non-empty string *s* (1 ≤ |*s*| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
#nhap du lieu vao, tao bien time = 0, t = 15, them vao vi tri a[0] so 0
#dung vong for duyet tu vi tri 1 den n+1 trong mang, if a[i] - a[i-1] < 15 thi time += a[i] -a[i-1], else thi time += t va print(time) sau do exit.
#tiep den la nhung truong hop khong nam trong vong for, if 90 - a[n] <= 15 thi time += 90 - a[n] else thi time += 15, sau do print(time)

n = int(input())
a = list(map(int,input().split()))
time = 0
t = 15
a.insert(0, 0)
for i in range(1,n+1):
    if a[i] - a[i-1] <= 15:
        time += a[i] - a[i-1]
    else:
        time += t
        print(time)
        exit()
if 90 - a[n] <= 15:
    time += 90 - a[n]
else:
    time += 15
print(time)
