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;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const long long INF = 1e18;
const long long mod = 1000000007;
long long dp[1050000];
string S;
vector<pair<int, int> > to[1050000];
long long ans;
long long TopZero, BottomZero;
vector<long long> v;
bool AllSame(string S) {
  for (int i = 0; i < S.size(); i++) {
    if (S[i] != S[0]) return false;
  }
  return true;
}
deque<pair<long long, long long> > deq;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> S;
  int N = S.size();
  if (AllSame(S)) {
    cout << N << "\n";
    return 0;
  }
  for (int i = 0;; i++) {
    if (S[i] == '0') {
      TopZero++;
    } else {
      break;
    }
  }
  S.erase(S.begin(), S.begin() + TopZero);
  while (S.back() == '0') {
    BottomZero++;
    S.pop_back();
  }
  long long tmpnum = 0;
  for (int i = 1; i < S.size(); i++) {
    if (S[i] == '0')
      tmpnum++;
    else {
      v.push_back(tmpnum);
      tmpnum = 0;
    }
  }
  N = v.size();
  for (int i = (int)v.size() - 1; i >= 0; i--) {
    while (!deq.empty() and deq.front().first <= v[i]) deq.pop_front();
    deq.push_front({v[i], i + 1});
    for (auto tmp : deq) {
      to[i].push_back(tmp);
    }
  }
  dp[0] = 1;
  for (int i = 0; i <= N; i++) {
    ans += dp[i];
    ans %= mod;
    for (int j = 0; j < to[i].size(); j++) {
      long long idx = to[i][j].second;
      long long factor;
      if (j == 0) {
        factor = to[i][j].first + 1;
      } else {
        factor = to[i][j].first - to[i][j - 1].first;
      }
      dp[idx] += dp[i] * factor;
      dp[idx] %= mod;
    }
  }
  ans = ans * (TopZero + 1) % mod;
  ans = ans * (BottomZero + 1) % mod;
  cout << ans << "\n";
  return 0;
}