Obsessive String
Hamed has recently found a string t and suddenly became quite fond of it. He spent several days trying to find all occurrences of t in other strings he had. Finally he became tired and started thinking about the following problem. Given a string s how many ways are there to extract k ≥ 1 non-overlapping substrings from it such that each of them contains string t as a substring? More formally, you need to calculate the number of ways to choose two sequences a1, a2, ..., ak and b1, b2, ..., bk satisfying the following requirements:  k ≥ 1          t is a substring of string saisai + 1... sbi (string s is considered as 1-indexed). As the number of ways can be rather large print it modulo 109 + 7.
Input consists of two lines containing strings s and t (1 ≤ |s|, |t| ≤ 105). Each string consists of lowercase Latin letters.
Print the answer in a single line.
(s, t) = (input(), input())
(n, m) = (len(t), len(s) + 1)
d = 1000000007
g = [1] * m
f = k = 0
for i in range(1, m):
    if s[i - n:i] == t:
        k = i
    if k:
        f = (f + g[k - n]) % d
    g[i] += (g[i - 1] + f) % d
print(f)
(s, t) = (input(), input())
(n, m) = (len(t), len(s))
d = 10 ** 9 + 7
p = [0] * (n + 1)
(i, j) = (0, 1)
while j < n:
    if t[i] == t[j]:
        j += 1
        i += 1
        p[j] = i
    elif i:
        i = p[i]
    else:
        j += 1
i = j = 0
f = [0] * (m + 1)
while j < m:
    if t[i] == s[j]:
        i += 1
        j += 1
        if i == n:
            i = p[i]
            f[j - n] = 1
    elif i:
        i = p[i]
    else:
        j += 1
s = [0] * (m + 1)
k = m
for j in range(m - 1, -1, -1):
    if f[j]:
        k = j
    if k < m:
        f[j] = (f[j + 1] + s[k + n] + m - k - n + 1) % d
        s[j] = (s[j + 1] + f[j]) % d
print(f[0])
(s, t) = (input(), input())
(n, m) = (len(t), len(s) + 1)
d = 1000000007
g = [1] * m
f = k = 0
for i in range(1, m):
    if s[i - n:i] == t:
        k = i
    if k:
        f = (f + g[k - n]) % d
    g[i] += (g[i - 1] + f) % d
print(f)