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 mod = 1e9 + 7;
inline int addmod(int x) { return x >= mod ? x - mod : x; }
int n, a[1000005], at, st[1000005], tp, lazy[1000005], f[1000005];
vector<int> val[1000005];
char s[1000005];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '1') {
      a[++at] = sum;
      sum = 0;
      continue;
    }
    sum++;
  }
  a[++at] = sum;
  if (at == 1) {
    printf("%d\n", n);
    return 0;
  }
  for (int i = at - 1; i; i--) {
    while (tp && a[st[tp]] < a[i]) {
      val[i].push_back(st[tp]);
      tp--;
    }
    st[++tp] = i;
  }
  f[1] = a[1] + 1;
  int ans = 0;
  for (int i = 1; i < at; i++) {
    if (tp > 1) {
      if (tp > 2) lazy[tp - 2] = addmod(lazy[tp - 2] + lazy[tp - 1]);
      f[st[tp - 1]] =
          addmod(f[st[tp - 1]] +
                 1ll * lazy[tp - 1] * (a[st[tp - 1]] - a[st[tp]]) % mod);
      lazy[tp - 1] = 0;
    }
    tp--;
    ans = addmod(ans + f[i]);
    int sz = val[i].size();
    for (int j = sz - 1; j >= 0; j--) st[++tp] = val[i][j];
    if (tp > 1) lazy[tp - 1] = addmod(lazy[tp - 1] + f[i]);
    if (tp) f[st[tp]] = addmod(f[st[tp]] + 1ll * f[i] * (a[st[tp]] + 1) % mod);
  }
  ans = 1ll * ans * (a[at] + 1) % mod;
  printf("%d\n", ans);
  return 0;
}
