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.
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
char s[1000010];
int sta[1000010], ans[1000010], len[1000010], n, top, stasum;
inline void insert(int id) {
  while (top && len[sta[top]] <= len[id]) {
    stasum = (stasum + 1ll * ans[sta[top]] * (mod - len[sta[top]] - 1)) % mod;
    stasum = (stasum + 1ll * ans[sta[top - 1]] * (len[sta[top]] + 1)) % mod;
    top--;
  }
  stasum = (stasum + 1ll * ans[id] * (len[id] + 1)) % mod;
  stasum = (stasum + 1ll * ans[sta[top]] * (mod - len[id] - 1)) % mod;
  sta[++top] = id;
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  int TTag = 1;
  while (s[n] == '0') TTag++, n--;
  if (n == 0) return cout << TTag - 1 << endl, 0;
  int Ans = 0;
  for (int i = n; i; i--) {
    int tmp = i;
    i--;
    while (s[i] == '0') i--;
    i++;
    len[tmp] = tmp - i, ans[tmp] = stasum + 1;
    insert(tmp);
    if (i == 1) Ans = 1ll * ans[tmp] * (len[tmp] + 1) % mod;
  }
  return cout << 1ll * Ans * TTag % mod << endl, 0;
}
