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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long inf = 1e15 + 7, mod = 1e9 + 7;
const long long max_n = 1e5 + 1;
long long n;
string s;
vector<pair<long long, long long> > a;
void solve() {
  cin >> s;
  n = s.size();
  long long cnt = 1;
  for (long long i = 1; i <= n; i++) {
    if (i == n || s[i] != s[i - 1]) {
      a.emplace_back(cnt, s[i - 1] - '0');
      cnt = 0;
    }
    cnt++;
  }
  n = a.size();
  vector<long long> dp(n + 1);
  dp[n] = 1;
  dp[n - 1] = a[n - 1].first;
  deque<long long> id;
  long long res = 0;
  long long id_temp = 0;
  for (long long i = n - 1; i >= 0; i--) {
    if (a[i].second == 0) {
      if (i == n - 1) id_temp = 1;
      dp[i] = dp[i + 1] * (a[i].first) % mod;
      while (id.size() > id_temp && a[id.front()].first <= a[i].first) {
        long long v = id.front();
        res -= a[v].first * (dp[v + 1]);
        res %= mod;
        id.pop_front();
        if (id.size() > id_temp) {
          res += (a[v].first) * (dp[id.front() + 1]);
          res %= mod;
        }
      }
      if (id.size() > id_temp) {
        res -= (a[i].first) * dp[id.front() + 1];
      }
      res += a[i].first * dp[i + 1];
      id.emplace_front(i);
      res %= mod;
    } else {
      dp[i] = ((res + 1) * a[i].first) % mod;
      if (i + 2 < n) dp[i] = (dp[i] + dp[i + 2]) % mod;
    }
  }
  if (n == 1) {
    cout << (dp[0] % mod + mod) % mod << endl;
    return;
  }
  long long q = dp[0];
  if (a[1].second == 1) {
    q += dp[1];
  }
  cout << (q % mod + mod) % mod << endl;
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  solve();
}
