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;
const ll MOD = 1e9 + 7;
struct node {
  int l, r;
  ll su, lr, la;
  node *left, *right;
  ll get() { return ((lr ? 0 : su) + la * (r - l + 1)) % MOD; }
  void merge() { su = (left->get() + right->get()) % MOD; }
  void push() {
    if (lr) {
      left->lr = 1;
      left->la = 0;
      right->lr = 1;
      right->la = 0;
    }
    left->la = (left->la + la) % MOD;
    right->la = (right->la + la) % MOD;
    lr = la = 0;
  }
  node(int l, int r) : l(l), r(r), su(0), la(0), lr(0) {
    if (l < r) {
      int m = (l + r) / 2;
      left = new node(l, m);
      right = new node(m + 1, r);
    }
  }
  void reset(int rl, int rr) {
    if (r < rl || rr < l) return;
    if (rl <= l && r <= rr) {
      lr = 1;
      la = 0;
      return;
    }
    push();
    left->reset(rl, rr);
    right->reset(rl, rr);
    merge();
  }
  void upd(int rl, int rr, ll v) {
    if (rl > rr || r < rl || rr < l) return;
    if (rl <= l && r <= rr) {
      la = (la + v) % MOD;
      return;
    }
    push();
    left->upd(rl, rr, v);
    right->upd(rl, rr, v);
    merge();
  }
  ll qry(int rl, int rr) {
    if (rr < l || r < rl) return 0;
    if (rl <= l && r <= rr) return get();
    push();
    merge();
    return (left->qry(rl, rr) + right->qry(rl, rr)) % MOD;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  int cnt1 = 0;
  for (auto c : s) cnt1 += (c == '1');
  if (cnt1 == 0) {
    cout << s.size() << '\n';
    return 0;
  }
  ll mul = 1;
  ll tmp = 1;
  while (s.back() == '0') {
    tmp++;
    s.pop_back();
  }
  mul = (mul * tmp) % MOD;
  reverse(s.begin(), s.end());
  tmp = 1;
  while (s.back() == '0') {
    tmp++;
    s.pop_back();
  }
  mul = (mul * tmp) % MOD;
  if (cnt1 == 1) {
    cout << mul << '\n';
    return 0;
  }
  vector<ll> dp(cnt1);
  dp[0] = 1;
  int ptr = 0;
  tmp = 0;
  ll pv = 0;
  int sz = s.size();
  node tree(0, sz);
  for (int i = 1; i < s.size(); i++) {
    if (s[i] == '0') {
      tmp++;
      continue;
    }
    ptr++;
    dp[ptr] = ((tmp + 1) * dp[ptr - 1]) % MOD;
    if (tmp > 0) dp[ptr] = (dp[ptr] + tree.qry(0, tmp - 1)) % MOD;
    tree.upd(tmp, sz, dp[ptr - 1]);
    tree.reset(0, tmp - 1);
    pv = tmp;
    tmp = 0;
  }
  ll res = 0;
  for (auto x : dp) res = (res + x) % MOD;
  cout << (res * mul) % MOD << '\n';
}