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;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while ((c < '0' || c > '9') && c ^ '-') c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - 48;
    c = getchar();
  }
  return x * f;
}
const int N = 1e6 + 10, mo = 1e9 + 7;
char c[N];
int n;
inline int qp(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = 1ll * ans * x % mo;
    x = 1ll * x * x % mo;
    y >>= 1;
  }
  return ans;
}
int dao[N], f[N], sum[N], g[N], top = 0;
inline int qumo(int x) { return x + ((x >> 31) & mo); }
int main() {
  scanf("%s", c + 1);
  n = strlen(c + 1);
  int las = 0;
  for (int i = 1; i <= n; i++)
    if (c[i] == '1') dao[++top] = i - las - 1, las = i;
  dao[++top] = n - las;
  if (top == 1) {
    cout << n << "\n";
    return 0;
  }
  f[1] = sum[1] = (dao[1] + 1);
  for (int i = 2; i < top; i++) {
    for (int j = 0; j <= dao[i]; j++) f[i] = qumo(f[i] + g[j] - mo);
    f[i] = (f[i] + 1ll * sum[i - 1] * (dao[i] + 1)) % mo;
    sum[i] = qumo(sum[i - 1] + f[i] - mo);
    for (int j = 0; j <= dao[i]; j++) g[j] = mo - sum[i - 1];
  }
  cout << 1ll * (dao[top] + 1) * sum[top - 1] % mo << "\n";
  return 0;
}
