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>
constexpr int P = 1000000007;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::string s;
  std::cin >> s;
  std::vector<int> p{-1};
  int n = s.length();
  for (int i = 0; i < n; ++i)
    if (s[i] == '1') p.push_back(i);
  p.push_back(n);
  std::vector<int> c;
  for (int i = 0; i + 1 < int(p.size()); ++i) c.push_back(p[i + 1] - p[i] - 1);
  if (c.size() == 1) {
    std::cout << n << "\n";
    return 0;
  }
  int ans = 1ll * (c[0] + 1) * (c.back() + 1) % P;
  c.pop_back();
  c.erase(c.begin());
  int m = c.size();
  std::vector<int> dp(m + 1), pre(m + 1);
  dp[0] = pre[0] = 1;
  std::vector<int> stk;
  for (int i = 0; i < m; ++i) {
    dp[i + 1] = 1ll * dp[i] * (c[i] + 1) % P;
    while (!stk.empty() && c[i] > c[stk.back()]) {
      int v = c[stk.back()];
      int sum = pre[stk.back()];
      stk.pop_back();
      if (!stk.empty()) sum = (sum - pre[stk.back()] + P) % P;
      dp[i + 1] = (dp[i + 1] + 1ll * sum * (c[i] - v)) % P;
    }
    pre[i + 1] = (pre[i] + dp[i + 1]) % P;
    stk.push_back(i);
  }
  std::cout << 1ll * ans * pre[m] % P << "\n";
  return 0;
}
