Problem: 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.

Solution (in PYTHON3):

Solution:
MOD = 1000000007
def main():
    s = input()
    n = len(s)
    zero = [0] * n
    for i in range(n):
        if s[i] == '0':
            zero[i] = zero[i-1] + 1 if i else 1
    nxt, dp = [0] * (n+2), [0] * (n+2)
    for i in range(n+1):
        nxt[i] = n
    for i in range(n-1, -1, -1):
        dp[i] = ((zero[i] <= zero[n-1]) + dp[nxt[0]] + dp[nxt[zero[i] + 1]]) % MOD
        nxt[zero[i]] = i
    ans = n if nxt[0] >= n else dp[nxt[0]] * (nxt[0] + 1) % MOD
    print(ans)

main()