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;
const int N = 1000005, mod = 1000000007;
char s[N];
int n, t[N], nxt[N][2], nxt0[N];
long long f[N], ans;
void inc(long long &x, long long y) {
  if ((x += y) >= mod) x -= mod;
}
void ins(int x, int y) {
  for (int i = x; i; i -= i & -i) t[i] = min(t[i], y);
}
int query(int x) {
  int rt = n + 1;
  for (int i = x; i <= n + 1; i += i & -i) rt = min(rt, t[i]);
  return rt;
}
void init() {
  nxt[n + 1][0] = nxt[n + 1][1] = n + 1;
  for (int i = n; i >= 1; --i)
    if (s[i + 1] == '0')
      nxt[i][0] = i + 1, nxt[i][1] = nxt[i + 1][1];
    else
      nxt[i][1] = i + 1, nxt[i][0] = nxt[i + 1][0];
  for (int i = 1; i <= n + 1; ++i) t[i] = n + 1;
  for (int i = n; i; --i) {
    if (s[i] == '1') continue;
    int j = i;
    while (j > 1 && s[j - 1] == s[j]) --j;
    int len = i - j + 1, x = query(len + 1);
    ins(len, j);
    if (x == n + 1)
      nxt0[i] = n + 1;
    else
      nxt0[i] = x + len;
    i = j;
  }
  if (s[1] == '0') inc(f[1], 1);
  for (int i = 1; i <= n; ++i)
    if (s[i] == '1') {
      inc(f[i], 1);
      break;
    }
}
int main() {
  scanf("%s", s + 1), n = strlen(s + 1);
  init();
  bool jdg = 0;
  for (int i = 1; i <= n - 1; ++i)
    if (s[i] == '0') {
      if (s[i + 1] == '0')
        inc(f[i + 1], f[i]);
      else if (jdg)
        inc(f[nxt0[i]], f[i]);
      inc(f[nxt[i][1]], f[i]);
    } else {
      jdg = 1;
      inc(f[nxt[i][0]], f[i]);
      inc(f[nxt[i][1]], f[i]);
    }
  int x = 1;
  for (int i = n; i >= 1; --i)
    if (s[i] == '0')
      ++x;
    else
      break;
  if (x == n + 1)
    printf("%d\n", n);
  else {
    for (int i = n; i >= 1; --i)
      if (s[i] == '1') inc(ans, f[i]);
    printf("%lld\n", ans * x % mod);
  }
  return 0;
}