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;
long long gi() {
  char cc = getchar();
  long long cn = 0, flus = 1;
  while (cc < '0' || cc > '9') {
    if (cc == '-') flus = -flus;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') cn = cn * 10 + cc - '0', cc = getchar();
  return cn * flus;
}
const long long N = 1e6 + 5;
const long long P = 1e9 + 7;
long long n, m, pre[N], w[N], dp[N], sum[N], a[N];
char s[N];
signed main() {
  cin >> (s + 1), n = strlen(s + 1);
  for (register long long i = (1); i <= (n); ++i) a[i] = s[i] - '0';
  long long ff = 0;
  for (register long long i = (1); i <= (n); ++i)
    (a[i] == 1) ? pre[++m] = ff, ff = 0 : ++ff;
  pre[++m] = ff;
  if (m == 1) {
    cout << n << endl;
    exit(0);
  }
  long long ans = (pre[1] + 1) * (pre[m] + 1) % P;
  for (register long long i = (0); i <= (n); ++i) w[i] = 1;
  sum[1] = dp[1] = 1;
  for (register long long i = 2; i < m; ++i) {
    long long L = w[0];
    for (register long long j = 0; j <= pre[i]; ++j)
      L = w[j], dp[i] = (dp[i] + sum[i - 1] - sum[L - 1] + P) % P;
    for (register long long j = 0; j <= pre[i]; ++j) w[j] = i;
    sum[i] = (sum[i - 1] + dp[i]) % P;
  }
  ans = ans * sum[m - 1] % P;
  cout << ans << endl;
  return 0;
}
