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;
const long long N = 1e6 + 5, mod = 1e9 + 7;
long long n, st, d[N], pos[N], nxt_1[N], nxt_d[N], f[N], ans;
char s[N];
signed main() {
  scanf("%s", s + 1), n = strlen(s + 1);
  for (long long i = 1; i <= n; i++)
    if (s[i] == '1') {
      st = i;
      break;
    }
  if (!st) printf("%lld\n", n), exit(0);
  for (long long i = 1; i <= n; i++) {
    if (s[i] == '0')
      d[i] = d[i - 1] + 1;
    else
      d[i] = 0;
  }
  for (long long i = 0; i <= n + 1; i++) pos[i] = n + 1;
  for (long long i = n; i >= 1; i--)
    nxt_1[i] = pos[0], nxt_d[i] = pos[d[i] + 1], pos[d[i]] = i;
  f[st] = st;
  for (long long i = st; i < n; i++) {
    if (nxt_1[i] != n + 1) f[nxt_1[i]] = (f[nxt_1[i]] + f[i]) % mod;
    if (nxt_d[i] != n + 1) f[nxt_d[i]] = (f[nxt_d[i]] + f[i]) % mod;
  }
  for (long long i = st; i <= n; i++)
    if (d[i] <= d[n]) ans = (ans + f[i]) % mod;
  printf("%lld\n", ans);
  return 0;
}