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 N = 1e6 + 10, p = 1e9 + 7;
map<int, int> mp;
char s[N];
inline void A(int &x, int y) { x = (x + y >= p ? x + y - p : x + y); }
inline void Append(int w) {
  int v1 = 0, v2 = 0;
  map<int, int>::iterator it;
  for (it = mp.begin(); it != mp.end() && it->first <= w; ++it) {
    A(v1, it->second);
    v2 = (v2 + 1ll * it->second * (w - it->first + 1)) % p;
  }
  mp.erase(mp.begin(), it);
  A(mp[w + 1], v1);
  mp[0] = v2;
}
int main() {
  scanf("%s", s + 1);
  int ln = 1, rn = strlen(s + 1);
  int time_l = 1, time_r = 1;
  while (s[ln] == '0') ++time_l, ++ln;
  if (ln > rn) {
    printf("%d\n", rn);
    return 0;
  }
  while (s[rn] == '0') ++time_r, --rn;
  mp[0] = 1;
  while (ln != rn) {
    ++ln;
    int w = 0;
    while (s[ln] == '0') ++ln, ++w;
    Append(w);
  }
  int ans = 0;
  for (auto x : mp) A(ans, x.second);
  printf("%lld\n", 1ll * ans * time_l % p * time_r % p);
  return 0;
}
