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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void pre() {}
void solve() {}
const long long mod = 1e9 + 7;
vector<int> v;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  pre();
  string s;
  cin >> s;
  int cnt = 0;
  for (auto &i : s) {
    if (i == '1') {
      v.push_back(cnt);
      cnt = 0;
    } else
      cnt++;
  }
  v.push_back(cnt);
  if ((int)(v).size() == 1) {
    cout << v[0];
  } else if ((int)(v).size() == 2) {
    cout << 1ll * (v[0] + 1) * (v[1] + 1) % mod;
  } else {
    vector<pair<long long, int>> g;
    g.push_back(make_pair(0, 1e9));
    long long ans = v[0] + 1;
    for (int i = 1; i <= ((int)(v).size() - 2); ++i) {
      int x = v[i], lst = -1;
      long long gg = 0;
      long long gans = ans;
      while (g.back().second < x) {
        gg += (gans - g.back().first) * (x - lst) % mod;
        gg %= mod;
        lst = g.back().second;
        gans = g.back().first;
        g.pop_back();
      }
      gg += (gans - g.back().first) * (x - lst) % mod;
      gg %= mod;
      g.push_back(make_pair(ans, x));
      ans += gg;
      ans %= mod;
    }
    ans = ans % mod;
    ans = ans * (v.back() + 1) % mod;
    cout << (ans + mod) % mod;
  }
  return 0;
}