Koa the Koala has a binary string s of length n. Koa can perform no more than n-1 (possibly zero) operations of the following form:

In one operation Koa selects positions i and i+1 for some i with 1 ≤ i < |s| and sets s_i to max(s_i, s_{i+1}). Then Koa deletes position i+1 from s (after the removal, the remaining parts are concatenated).

Note that after every operation the length of s decreases by 1.

How many different binary strings can Koa obtain by doing no more than n-1 (possibly zero) operations modulo 10^9+7 (1000000007)?

Input

The only line of input contains binary string s (1 ≤ |s| ≤ 10^6). For all i (1 ≤ i ≤ |s|) s_i = 0 or s_i = 1.

Output

On a single line print the answer to the problem modulo 10^9+7 (1000000007).

Examples

Input


000


Output


3


Input


0101


Output


6


Input


0001111


Output


16


Input


00101100011100


Output


477

Note

In the first sample Koa can obtain binary strings: 0, 00 and 000.

In the second sample Koa can obtain binary strings: 1, 01, 11, 011, 101 and 0101. For example:

  * to obtain 01 from 0101 Koa can operate as follows: 0101 → 0(10)1 → 011 → 0(11) → 01. 
  * to obtain 11 from 0101 Koa can operate as follows: 0101 → (01)01 → 101 → 1(01) → 11. 



Parentheses denote the two positions Koa selected in each operation.
import sys

MOD = 1000000007


def red(x):
    return x if x < MOD else x - MOD


S = input()
n = len(S)

i = 0
while i < n and S[i] == '0':
    i += 1
if i == n:
    print (n)
    sys.exit()
j = n
while j and S[j - 1] == '0':
    j -= 1

fac = (i + 1) * (n - j + 1)
S = S[i:j]
n = len(S)

count0 = [+(c == '0') for c in S]
for i in reversed(range(n - 1)):
    if count0[i]:
        count0[i] += count0[i + 1]

ways0 = 1
ways1 = 0

buckets = [0] * (n + 1)
buckets[0] = 1
for i in reversed(range(n)):
    if count0[i]:
        c = count0[i]
        ways0 = red(ways0 - buckets[c] + MOD)

        buckets[c] = ways1
        ways0 = red(ways0 + ways1)
    else:
        ways1 = red(ways1 + ways0)
print (ways1 * fac % MOD)