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:
import sys
readline = sys.stdin.readline
MOD = 10**9+7;S = readline().strip().split('1')
if len(S) == 1:print(len(S[0]))
else:
    S = [len(s)+1 for s in S];ans = S[0]*S[-1];S = S[1:-1];dp = [0]*(max(S)+2);dp[0] = 1
    for ai in S:
        res = 0;rz = 0
        for i in range(ai+1):res = (res + dp[i])%MOD;rz = (rz + (ai-i)*dp[i])%MOD;dp[i] = 0
        dp[0] = rz;dp[ai] = res
    aaa = 0
    for d in dp:aaa = (aaa+d)%MOD
    print(aaa*ans%MOD)