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;
string s;
int to[2100000][2];
int nxt[2100000][2];
long long dp[2100000];
int main() {
  cin >> s;
  int lst1 = 0x3f3f3f3f;
  nxt[s.size()][1] = nxt[s.size()][0] = 0x3f3f3f3f;
  for (int i = s.size() - 1; i >= 0; i--) {
    nxt[i][s[i] - '0'] = i;
    nxt[i][(s[i] - '0') ^ 1] = nxt[i + 1][(s[i] - '0') ^ 1];
  }
  int cnt0 = 0;
  set<pair<int, int> > st;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '0') cnt0++;
    to[i][1] = nxt[i + 1][1];
    if (s[i] == '1')
      to[i][0] = nxt[i][0];
    else {
      if (s[i + 1] == '0')
        to[i][0] = i + 1;
      else {
        to[i][0] = 0x3f3f3f3f;
        while (!st.empty() && cnt0 >= st.begin()->first) {
          to[st.begin()->second][0] = i - cnt0 + st.begin()->first;
          st.erase(st.begin());
        }
        st.insert({cnt0 + 1, i});
      }
    }
    if (s[i] == '1') cnt0 = 0;
  }
  long long ans = 0;
  long long pos = 0;
  while (s[pos] != '1' && pos < s.size()) pos++;
  if (pos == s.size()) {
    cout << s.size();
    return 0;
  }
  int lst = s.size() - 1;
  while (s[lst] != '1') lst--;
  dp[pos] = 1;
  for (int i = pos; i < s.size(); i++) {
    dp[i] %= 1000000007;
    if (s[i] == '1') ans += dp[i];
    if (to[i][0] < s.size()) dp[to[i][0]] += dp[i];
    if (to[i][1] < s.size()) dp[to[i][1]] += dp[i];
  }
  pos = (pos + 1) * (s.size() - lst) % 1000000007;
  cout << pos * ans % 1000000007;
  return ~~(0 ^ 0 ^ 0);
}