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>
template <class T>
T read(T &a) {
  a = 0;
  char x = getchar();
  bool f = 0;
  for (; x < '0' || x > '9'; x = getchar()) f |= x == '-';
  for (; x >= '0' && x <= '9'; x = getchar()) a = (a << 3) + (a << 1) + x - '0';
  if (f) a = -a;
  return a;
}
using namespace std;
const int N = 1e6 + 5, MOD = 1e9 + 7;
char s[N];
int pre[N], suf[N], nxt[N], pos[N];
long long f[N];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  int s_pos = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '1') {
      s_pos = i;
      break;
    }
  }
  if (!s_pos) {
    return printf("%d\n", n), 0;
  }
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '0') {
      pre[i] = pre[i - 1] + 1;
    } else
      pre[i] = 0;
  }
  for (int i = 0; i <= n + 1; ++i) {
    pos[i] = n + 1;
  }
  for (int i = n; i > 0; --i) {
    suf[i] = pos[0];
    nxt[i] = pos[pre[i] + 1];
    pos[pre[i]] = i;
  }
  f[s_pos] = s_pos;
  for (int i = s_pos; i < n; ++i) {
    if (suf[i] != n + 1) {
      f[suf[i]] = (f[suf[i]] + f[i]) % MOD;
    }
    if (nxt[i] != n + 1) {
      f[nxt[i]] = (f[nxt[i]] + f[i]) % MOD;
    }
  }
  long long ans = 0;
  for (int i = s_pos; i <= n; ++i) {
    if (pre[i] <= pre[n]) {
      ans += f[i];
      ans %= MOD;
    }
  }
  cout << ans << "\n";
  return 0;
}