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;
namespace strangeoperation {
long long t;
long long mod = 1000000007;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  long long n = s.length();
  long long a[n];
  long long num1 = 0;
  for (long long i = 0; i < n; i++) {
    a[i] = s[i] - '0';
    if (a[i] == 1) {
      num1++;
    }
  }
  if (num1 == 0) {
    cout << n << endl;
    return 0;
  }
  if (num1 == 1) {
    long long part1 = 0;
    long long part2 = 0;
    for (long long i = 0; i < n; i++) {
      if (a[i] == 1) {
        break;
      }
      part1++;
    }
    for (long long i = n - 1; i >= 0; i--) {
      if (a[i] == 1) {
        break;
      }
      part2++;
    }
    long long ans = (1 + part1) * (1 + part2);
    cout << (ans % mod) << endl;
    return 0;
  }
  vector<long long> p;
  long long cnt = 0;
  long long x = num1;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 1) {
      p.push_back(cnt + 1);
      cnt = 0;
      continue;
    }
    cnt++;
  }
  p.push_back(cnt + 1);
  long long prev[x + 1];
  prev[1] = -1;
  stack<long long> st;
  st.push(1);
  for (long long i = 2; i <= x; i++) {
    long long idx;
    while (!st.empty()) {
      idx = st.top();
      if (p[idx] > p[i]) {
        prev[i] = idx;
        st.push(i);
        break;
      }
      st.pop();
    }
    if (st.empty()) {
      prev[i] = -1;
      st.push(i);
    }
  }
  long long dp[x + 1];
  dp[0] = 1;
  dp[1] = p[1] + 1;
  for (long long i = 2; i < x; i++) {
    long long j = prev[i];
    if (j == -1) {
      dp[i] = 1 + p[i] * dp[i - 1];
      dp[i] %= mod;
      continue;
    }
    dp[i] = 1 + p[i] * dp[i - 1] + dp[j] - p[i] * dp[j - 1] - 1;
    dp[i] %= mod;
  }
  long long ans = (dp[x - 1] % mod + mod) % mod;
  ans *= p[0];
  ans %= mod;
  ans *= p[x];
  ans %= mod;
  cout << (mod + ans) % mod << endl;
  return 0;
}
}  // namespace strangeoperation
int main() {
  strangeoperation::main();
  return 0;
}