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;
using ll = long long;
const int MOD = 1e9 + 7;
void add(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  string S;
  cin >> S;
  int tail = 0;
  while (S.size() && S.back() == '0') {
    tail++;
    S.pop_back();
  }
  if (S.empty()) {
    cout << tail << "\n";
    return 0;
  }
  reverse(S.begin(), S.end());
  int head = 0;
  while (S.size() && S.back() == '0') {
    head++;
    S.pop_back();
  }
  reverse(S.begin(), S.end());
  int N = S.size();
  S = '~' + S;
  vector<int> nxt0(N + 1);
  vector<int> nxt1(N + 1);
  {
    int last = N + 1;
    for (int i = N; i >= 0; --i) {
      nxt1[i] = last;
      if (S[i] == '1') last = i;
    }
  }
  {
    vector<int> go(N + 1);
    for (int i = 0; i <= N; ++i) {
      if (S[i] != '0')
        go[i] = 0;
      else
        go[i] = go[i - 1] + 1;
    }
    vector<int> last(N + 2, N + 1);
    for (int i = N; i >= 0; --i) {
      nxt0[i] = last[go[i] + 1];
      last[go[i]] = i;
    }
  }
  vector<int> dp(N + 1);
  dp[1] = 1;
  for (int i = 1; i <= N; ++i) {
    if (nxt0[i] != N + 1) add(dp[nxt0[i]], dp[i]);
    if (nxt1[i] != N + 1) add(dp[nxt1[i]], dp[i]);
  }
  int ans = 0;
  for (int i = 1; i <= N; ++i)
    if (S[i] == '1') add(ans, dp[i]);
  ans = ll(ans) * (head + 1) % MOD;
  ans = ll(ans) * (tail + 1) % MOD;
  cout << ans << "\n";
  return 0;
}