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>
char ss[1001000];
int nm[1001000][2], ta, f[1001000], g[1001000], s[1001000], q[1001000];
const int mod = 1e9 + 7;
int main() {
  register int n;
  register char la = '1';
  ta = 1;
  nm[1][0] = 1 << 30;
  scanf("%s", ss + 1);
  n = strlen(ss + 1);
  for (register int i = 1; i <= n; i++) {
    if (la > ss[i]) ta++;
    la = ss[i];
    nm[ta][la - '0']++;
  }
  n = ta;
  q[ta = 1] = 1;
  s[1] = g[1] = nm[1][1];
  if (!nm[1][1]) g[1]++;
  register int nn = 1;
  for (register int i = 2; i <= n; i++) {
    f[i] = 1ll * nm[i][0] * g[i - 1] % mod;
    while (nm[i][0] > nm[q[ta]][0]) {
      f[i] = (f[i] + 1ll * (nm[i][0] - nm[q[ta]][0]) *
                         (s[q[ta] - 1] - s[q[ta - 1] - 1] + mod)) %
             mod;
      ta--;
    }
    q[++ta] = i;
    nn = (nn + f[i]) % mod;
    g[i] = 1ll * nn * nm[i][1] % mod;
    s[i] = (s[i - 1] + g[i]) % mod;
  }
  register int ans = 0;
  for (register int i = 1 + (!nm[1][1]); i <= n; i++) ans = (ans + g[i]) % mod;
  if (nm[n][1] == 0) {
    for (register int i = 1 + ((!nm[1][1]) && nm[n][0] != strlen(ss + 1));
         i < n; i++)
      ans = (ans + 1ll * g[i] * nm[n][0]) % mod;
  }
  printf("%d\n", ans);
}
