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 long long MOD = 1e9 + 7;
const int SIZE = 1e6 + 5;
const long long INF = 1LL << 58;
const double eps = 1e-13;
string s;
int dist[1000009];
int nxt[1000009];
long long dp[1000009];
int main() {
  cin >> s;
  int suf0 = 0;
  while (((int)(s).size()) && s.back() == '0') {
    s.pop_back();
    suf0++;
  }
  if (s.empty()) {
    cout << suf0;
    return 0;
  }
  int n = ((int)(s).size());
  for (int i = 0; i < (n); ++i) {
    if (s[i] == '0') dist[i] = (i ? dist[i - 1] : 0) + 1;
  }
  for (int i = 0; i < (n); ++i) nxt[i] = n;
  for (int i = n - 1; i >= 0; i--) {
    dp[i] = (dp[i] + (s[i] == '1') + dp[nxt[0]] + dp[nxt[dist[i] + 1]]) % MOD;
    nxt[dist[i]] = i;
  }
  long long ans = n;
  if (nxt[0] < n) {
    ans = (dp[nxt[0]] * (nxt[0] + 1)) % MOD;
  }
  ans = (ans * (suf0 + 1)) % MOD;
  printf("%lld", ans);
}