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.
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long dp[1000009];
long long s[1000009];
int main() {
  string str;
  cin >> str;
  long long c = str[0], k = 1;
  long long mlt = 1;
  vector<pair<long long, long long> > v;
  for (long long i = 1; i < str.size(); i++) {
    if (str[i] == c) {
      k++;
      continue;
    } else {
      if (v.size() == 0 && c == '0') {
        mlt = k + 1;
      } else
        v.push_back({k, c - '0'});
      k = 1;
      c = str[i];
    }
  }
  v.push_back({k, c - '0'});
  if (v.size() == 1) {
    cout << v[0].first * mlt % mod << endl;
    return 0;
  }
  long long sz = 1;
  long long ans = 0;
  stack<pair<long long, long long> > S;
  for (long long i = 0; i < v.size(); i++) {
    if (v[i].second == 1) {
      dp[i] = sz * v[i].first % mod;
      s[i] = (s[i - 1] + dp[i]) % mod;
    } else {
      if (i) s[i] = s[i - 1];
      long long u = v[i].first;
      long long l = i, z = 0;
      while (S.size()) {
        long long L = S.top().first, j = S.top().second;
        dp[i] = ((dp[i] + (s[l] - s[j]) * (u - z)) % mod + mod) % mod;
        z = L;
        l = j;
        if (L < u)
          S.pop();
        else
          break;
      }
      if (!S.size()) dp[i] = ((dp[i] + s[l] * (u - z)) % mod + mod) % mod;
      S.push({v[i].first, i});
      sz = (sz + dp[i]) % mod;
    }
  }
  if (v.back().second == 1)
    cout << s[v.size() - 1] * mlt % mod << endl;
  else
    cout << (s[v.size() - 1]) * (v.back().first + 1) % mod * mlt % mod << endl;
}
