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;
const int MAXN = 1e6 + 5, Mod = 1e9 + 7;
char buf[1 << 12], *pp1 = buf, *pp2 = buf, nc;
int ny;
inline char gc() { return getchar(); }
inline int read() {
  int x = 0;
  for (ny = 1; nc = gc(), (nc < 48 || nc > 57) && nc != EOF;)
    if (nc == 45) ny = -1;
  if (nc < 0) return nc;
  for (x = nc - 48; nc = gc(), 47 < nc && nc < 58 && nc != EOF;
       x = (x << 3) + (x << 1) + (nc ^ 48))
    ;
  return x * ny;
}
int n, p[MAXN], sum[MAXN], cnt, last[MAXN], f[MAXN];
char s[MAXN];
int main() {
  scanf("%s", s + 1), n = strlen(s + 1);
  int t = 0;
  for (int i = (1); i <= (n); i++)
    if (s[i] == '1')
      p[++cnt] = t, t = 0;
    else
      t++;
  p[++cnt] = t;
  if (cnt == 1) {
    cout << n << '\n';
    return 0;
  }
  int ans = 1ll * (p[1] + 1) * (p[cnt] + 1) % Mod;
  for (int i = (0); i <= (n); i++) last[i] = 1;
  sum[1] = 1;
  for (int i = (2); i <= (cnt - 1); i++) {
    for (int j = (0); j <= (p[i]); j++)
      f[i] = (0ll + f[i] + sum[i - 1] - sum[last[j] - 1] + Mod) % Mod;
    for (int j = (0); j <= (p[i]); j++) last[j] = i;
    sum[i] = (sum[i - 1] + f[i]) % Mod;
  }
  cout << 1ll * ans * (sum[cnt - 1]) % Mod;
  return 0;
}
