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>
int f[1010000], s[1010000];
char c[1010000];
int a[1010000], p[1010000], nxt1[1010000];
int st[1010000], top = 0;
int nxt0[1010000];
int main() {
  scanf("%s", c + 1);
  bool ok = 0;
  int n = strlen(c + 1);
  int l = n, r = 1;
  for (int i = 1; i <= n; i++) {
    if (c[i] == '1') {
      ok = 1;
      l = std::min(l, i);
      r = std::max(r, i);
    }
  }
  if (!ok) return 0 * printf("%d\n", n);
  if (l == r) return 0 * printf("%lld\n", 1ll * l * (n - r + 1) % 1000000007);
  int N = 0;
  for (int i = l; i <= r; i++) {
    if (c[i] == '0') {
      if (c[i] == c[i - 1])
        a[N]++;
      else
        a[++N] = 1, p[N] = i;
    }
  }
  for (int i = n; i >= 1; i--) {
    nxt0[i] = nxt0[i + 1];
    nxt1[i] = nxt1[i + 1];
    if (c[i + 1] == '0')
      nxt0[i] = i + 1;
    else
      nxt1[i] = i + 1;
  }
  for (int i = 1; i <= N; i++) {
    while (top && a[i] > a[st[top]]) {
      nxt0[p[st[top]] + a[st[top]] - 1] = p[i] + a[st[top]];
      top--;
    }
    st[++top] = i;
  }
  for (int i = 1; i <= top; i++) nxt0[p[st[i]] + a[st[i]] - 1] = 0;
  l++;
  r--;
  f[l - 1] = 1;
  for (int i = l - 1; i < r; i++) {
    if (nxt1[i]) f[nxt1[i]] = (f[nxt1[i]] + f[i]) % 1000000007;
    if (nxt0[i]) f[nxt0[i]] = (f[nxt0[i]] + f[i]) % 1000000007;
  }
  int sum = 1;
  for (int i = l - 1; i <= r; i++) sum = (sum + f[i]) % 1000000007;
  printf("%lld\n", 1ll * sum * (l - 1) % 1000000007 * (n - r) % 1000000007);
}
