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;
const int MOD = (int)1e9 + 7;
const int N = (int)1e6 + 10;
int dp[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  string s;
  cin >> s;
  int n = s.size();
  bool sol = false;
  for (auto x : s) {
    if (x == '1') sol = true;
  }
  if (!sol) {
    cout << n << "\n";
    return 0;
  }
  deque<int> solv;
  for (int i = 0; i < s.size(); i++) {
    solv.push_back(s[i] - '0');
  }
  int pf = 0;
  while (solv.front() == 0) {
    pf++;
    solv.pop_front();
  }
  int sf = 0;
  while (solv.back() == 0) {
    sf++;
    solv.pop_back();
  }
  int aa = 0;
  vector<int> seg;
  for (int i = 1; i < solv.size(); i++) {
    if (solv[i] == 1) {
      seg.push_back(aa);
      aa = 0;
    } else {
      aa++;
    }
  }
  int outp = 1;
  int di;
  int nw;
  for (auto x : seg) {
    nw = outp;
    for (int j = 0; j <= x; j++) {
      di = (dp[j] + (outp - dp[j] + MOD) % MOD) % MOD;
      nw = (nw + (di - dp[j] + MOD) % MOD) % MOD;
      dp[j] = di;
    }
    outp = nw;
  }
  outp = (outp * 1ll * (pf + 1)) % MOD;
  outp = (outp * 1ll * (sf + 1)) % MOD;
  cout << outp << "\n";
  return 0;
}
