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 maxn = 1e6 + 10, mod = 1e9 + 7, inf = 1e9 + 10;
int dp[maxn], SM[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie();
  string s;
  cin >> s;
  int cnt = 0, cnt2 = 0;
  for (int i = 0; i < int((s).size()); i++) {
    if (s[i] == '1') break;
    cnt++;
  }
  if (cnt == int((s).size())) return cout << cnt << endl, 0;
  s = s.substr(cnt);
  reverse(s.begin(), s.end());
  for (int i = 0; i < int((s).size()); i++) {
    if (s[i] == '1') break;
    cnt2++;
  }
  s = s.substr(cnt2);
  reverse(s.begin(), s.end());
  s.pop_back();
  if (s.empty()) return cout << 1ll * (cnt + 1) * (cnt2 + 1) % mod, 0;
  int lst = 0;
  dp[0] = 1;
  SM[1] = 1;
  for (int i = 1; i < int((s).size()); i++) {
    if (s[i] == '1')
      lst = 0;
    else
      lst++;
    SM[i + 1] = (SM[i] - dp[lst] + SM[i - lst] + 1) % mod;
    SM[i + 1] = (SM[i + 1] + mod) % mod;
    dp[lst] = SM[i - lst] + 1;
  }
  return cout << 1ll * (SM[int((s).size())] + 1) * (cnt + 1) % mod *
                     (cnt2 + 1) % mod
              << endl,
         0;
}
