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.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int cys = 1000000007;
int n;
int a[1000005], nxt[1000005], pl[1000005];
long long d[1000005];
vector<int> vec[1000005];
char s[1000005];
long long mod(long long x) { return x >= cys ? x - cys : x; }
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; i++) {
    if (s[i] == '1')
      a[i] = 0;
    else
      a[i] = a[i - 1] + 1;
  }
  for (int i = 1; i <= n; i++) vec[a[i]].push_back(i);
  int now = n + 1, lst = 0;
  for (int i = n; i >= 0; i--) {
    nxt[i] = now;
    if (s[i] == '1') now = i, chkmax(lst, i);
  }
  if (now == n + 1) return printf("%d\n", n), 0;
  d[0] = 1;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (i) pl[a[i]]++;
    if (s[i + 1] == '1') {
      d[i + 1] = mod(d[i + 1] + d[i]);
      if (pl[a[i] + 1] < vec[a[i] + 1].size() && i >= now) {
        d[vec[a[i] + 1][pl[a[i] + 1]]] =
            mod(d[vec[a[i] + 1][pl[a[i] + 1]]] + d[i]);
      }
    } else {
      d[i + 1] = mod(d[i + 1] + d[i]);
      if (nxt[i] <= n) d[nxt[i]] = mod(d[nxt[i]] + d[i]);
    }
    if (a[i] <= a[n] && i >= now) ans = mod(ans + d[i]);
  }
  printf("%lld\n", mod(ans + d[n]));
  return 0;
}
