Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, a[1000005];
char s[1000005];
int ans;
int suf0[1000005], nxt1[1000005], nxt0s[1000005], stk[1000005], tp;
int f[1000005];
int main() {
  scanf("%s", s + 1), n = strlen(s + 1);
  for (int i = 1; i <= (n); ++i) a[i] = s[i] - '0';
  int pre = 0, suf = 0;
  while (pre < n && !a[pre + 1]) ++pre;
  while (suf < n && !a[n - suf]) ++suf;
  if (pre == n && suf == n) return printf("%d\n", n), 0;
  if (pre + suf == n - 1)
    return printf("%lld\n", (long long)(pre + 1) * (suf + 1) % 1000000007), 0;
  for (int i = 1; i <= (n - pre - suf - 2); ++i) a[i] = a[pre + 1 + i];
  n -= pre + suf + 2;
  suf0[0] = 0;
  for (int i = 1; i <= (n); ++i) suf0[i] = a[i] ? 0 : suf0[i - 1] + 1;
  nxt1[n] = n + 1;
  for (int i = n - 1; i >= (0); --i) nxt1[i] = a[i + 1] ? i + 1 : nxt1[i + 1];
  for (int i = n; i >= (0); --i) {
    while (tp && suf0[stk[tp]] <= suf0[i]) --tp;
    if (tp)
      nxt0s[i] = stk[tp];
    else
      nxt0s[i] = n + 1;
    stk[++tp] = i;
  }
  f[0] = 1;
  for (int i = 0; i <= (n - 1); ++i) {
    if (nxt1[i] <= n) f[nxt1[i]] = (f[nxt1[i]] + f[i]) % 1000000007;
    if (nxt0s[i] <= n) f[nxt0s[i]] = (f[nxt0s[i]] + f[i]) % 1000000007;
  }
  ans = 1;
  for (int i = 0; i <= (n); ++i) ans = (ans + f[i]) % 1000000007;
  printf("%lld\n",
         (long long)ans * (pre + 1) % 1000000007 * (suf + 1) % 1000000007);
  return 0;
}