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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ul = unsigned long long;
using ld = long double;
const int N = 1 << 20, MOD = 1e9 + 7;
void add(ll &x, ll y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  ll n = s.size(), z = 0, fz = 0;
  vector<pair<ll, ll>> li;
  ll ans = 0;
  for (int i = 0; i < (n); i++) {
    z++;
    if (s[i] == '1') {
      if (fz == 0) {
        fz = z;
        z = 0;
        continue;
      }
      ll ca = (ans + 1) % MOD, cz = z;
      ans = (ans + ca * z) % MOD;
      while (!li.empty() && z > 0) {
        if (li.back().first <= z) {
          ans = (ans + MOD - li.back().second * li.back().first % MOD) % MOD;
          z -= li.back().first;
          li.pop_back();
        } else {
          ans = (ans + MOD - li.back().second * z % MOD) % MOD;
          li.back().first -= z;
          z = 0;
        }
      }
      li.push_back({cz, ca});
      z = 0;
    }
  }
  ans = (ans + 1) * fz % MOD * (z + 1) % MOD;
  if (!fz) ans = n;
  cout << ans;
}