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;
template <class T>
int chkmax(T &a, T b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
int chkmin(T &a, T b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class iterator>
void output(iterator begin, iterator end, ostream &out = cerr) {
  while (begin != end) {
    out << (*begin) << " ";
    begin++;
  }
  out << endl;
}
template <class T>
void output(T x, ostream &out = cerr) {
  output(x.begin(), x.end(), out);
}
void fast_io() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int MOD = 1e9 + 7;
int mul(int a, int b) { return (1LL * a * b) % MOD; }
void add(int &a, int b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
}
void subt(int &a, int b) {
  a -= b;
  if (a < 0) {
    a += MOD;
  }
}
int n, ones_sum, zeros_sum, head = -1, pos, first_zero;
string s;
vector<int> zeros, nxt;
void add0() {
  if (first_zero) {
    pos = zeros.size();
    first_zero = 0;
  }
  add(zeros_sum, ones_sum);
  zeros.push_back(ones_sum);
  nxt.push_back(head);
  head = (int)zeros.size() - 1;
  if (nxt[pos] != -1) {
    subt(zeros_sum, zeros[nxt[pos]]);
    nxt[pos] = nxt[nxt[pos]];
  }
}
void add1() {
  add(ones_sum, zeros_sum + 1);
  first_zero = 1;
}
int mult = 1, mult2 = 1;
void read() {
  cin >> s;
  n = s.size();
  int pref_len = 0, suf_len = 0;
  while (pref_len < n && s[pref_len] == '0') {
    pref_len++;
    mult++;
  }
  while (suf_len < n && s[n - 1 - suf_len] == '0') {
    suf_len++;
    mult2++;
  }
  if (pref_len == n) {
    cout << n << '\n';
    exit(0);
  }
  s.substr(pref_len, n - suf_len - pref_len);
  n = s.size();
}
void solve() {
  for (int i = 0; i < n; ++i) {
    if (s[i] == '1') {
      add1();
    } else {
      add0();
    }
  }
}
void calc_ans() {
  int ans = ones_sum;
  ans = mul(ans, mult);
  ans = mul(ans, mult2);
  cout << ans << '\n';
}
signed main() {
  fast_io();
  read();
  solve();
  calc_ans();
}