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;
constexpr long double m_pi = 3.1415926535897932L;
constexpr long long MOD = 1000000007;
constexpr long long INF = 1LL << 61;
constexpr long double EPS = 1e-10;
template <typename T>
using vector2 = vector<vector<T>>;
template <typename T>
using vector3 = vector<vector2<T>>;
long long dp[1010101];
int main() {
  ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  string s;
  cin >> s;
  long long n = s.size();
  vector<long long> cnt(n + 1, 0);
  for (int i = n - 1; i >= 0; i--)
    if (s[i] == '0') cnt[i] = (i == n - 1 ? 0 : cnt[i + 1]) + 1;
  vector<long long> last(n + 2, -1);
  vector<long long> num(n, 0);
  for (long long(i) = (long long)(0); i < (long long)(n); i++)
    if (s[i] == '0') num[i] = (i == 0 ? 0 : num[i - 1]) + 1;
  vector<long long> nxt(n, -1);
  for (int i = n - 1; i >= 0; i--) {
    nxt[i] = last[num[i] + 1];
    last[num[i]] = i;
  }
  dp[cnt[0]] = 1;
  for (long long(i) = (long long)(0); i < (long long)(n); i++) {
    (dp[i + 1 + cnt[i + 1]] += dp[i]) %= MOD;
    if (nxt[i] != -1) (dp[nxt[i]] += dp[i]) %= MOD;
  }
  if (cnt[0] == n) {
    cout << n << "\n";
    return 0;
  }
  long long ans = 0;
  long long mx = num[n - 1];
  for (long long(i) = (long long)(cnt[0]); i < (long long)(n); i++) {
    if (s[i] != '0')
      (ans += dp[i]) %= MOD;
    else if (num[i] <= mx)
      (ans += dp[i]) %= MOD;
  }
  cout << ans * (cnt[0] + 1) % MOD << "\n";
  return 0;
}