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;
double EPS = 1e-9;
int INF = 1000000005;
long long INFF = 1000000000000000005LL;
double PI = acos(-1);
int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
inline string IntToString(long long a) {
  char x[100];
  sprintf(x, "%lld", a);
  string s = x;
  return s;
}
inline long long StringToInt(string a) {
  char x[100];
  long long res;
  strcpy(x, a.c_str());
  sscanf(x, "%lld", &res);
  return res;
}
inline string GetString(void) {
  char x[1000005];
  scanf("%s", x);
  string s = x;
  return s;
}
inline string uppercase(string s) {
  int n = (int)s.size();
  for (int(i) = (0); (i) < (n); ++(i))
    if (s[i] >= 'a' && s[i] <= 'z') s[i] = s[i] - 'a' + 'A';
  return s;
}
inline string lowercase(string s) {
  int n = (int)s.size();
  for (int(i) = (0); (i) < (n); ++(i))
    if (s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a';
  return s;
}
inline void OPEN(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
set<string> risan;
void dfs(string s) {
  risan.insert(s);
  for (int(i) = (0); (i) < ((int)s.size() - 1); ++(i)) {
    string t = "";
    for (int(j) = (0); (j) < ((int)s.size()); ++(j)) {
      if (j == i)
        t += max(s[i], s[i + 1]);
      else if (j == i + 1)
        ;
      else
        t += s[j];
    }
    dfs(t);
  }
}
long long MOD = 1000000007;
int ada[10];
int ada2[10];
int tambah[1000005];
long long lst = 0;
long long add = 0;
int main() {
  string s;
  s = GetString();
  if (s == string((int)s.size(), '0')) {
    printf("%d\n", (int)s.size());
    return 0;
  }
  int a = 0;
  while (s.back() == '0') {
    s.pop_back();
    ++a;
  }
  reverse(s.begin(), s.end());
  int b = 0;
  while (s.back() == '0') {
    s.pop_back();
    ++b;
  }
  reverse(s.begin(), s.end());
  ;
  int nol = 0;
  for (int(i) = (0); (i) < ((int)s.size()); ++(i)) {
    if (s[i] == '1') {
      nol = 0;
      lst += add + 1;
    } else {
      ++nol;
      add -= tambah[nol];
      tambah[nol] = lst;
      add += tambah[nol];
    }
    lst %= MOD;
    add %= MOD;
    lst = (lst % MOD + MOD) % MOD;
    add = (add % MOD + MOD) % MOD;
  }
  lst = (lst % MOD + MOD) % MOD;
  lst = (lst * (a + 1)) % MOD;
  lst = (lst * (b + 1)) % MOD;
  printf("%lld\n", lst);
  return 0;
}