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 long long sz = 2e5 + 5, mod = 1e9 + 7;
long long t, n, m, k, q;
const long long inf = 3e18;
long long dx[] = {1, 0, -1, 0};
long long dy[] = {0, 1, 0, -1};
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  t = 1;
  while (t--) {
    string s;
    cin >> s;
    long long n = (long long)s.size();
    long long beg = 0, end = n;
    while (beg < end && s[beg] == '0') {
      ++beg;
    }
    while (beg < end && s[end - 1] == '0') {
      --end;
    }
    if (beg == end) {
      cout << n << '\n';
      return 0;
    }
    vector<long long> aux(n + 1);
    long long sum_aux = 0;
    vector<long long> dp(n);
    long long ones = 0;
    long long zeros = 0;
    long long sum_for_ones = 0;
    for (long long i = end - 1; i >= beg; i--) {
      if (s[i] == '0') {
        ++zeros;
        sum_aux -= aux[zeros];
        sum_aux += mod;
        sum_aux %= mod;
        aux[zeros] = dp[i + zeros];
        sum_aux += aux[zeros];
        sum_aux %= mod;
        dp[i] = dp[i + 1];
        continue;
      }
      sum_for_ones += sum_aux;
      sum_for_ones %= mod;
      zeros = 0;
      ++ones;
      dp[i] = ones + sum_for_ones;
      dp[i] %= mod;
    }
    long long ans = dp[beg] * (beg + 1);
    ans %= mod;
    ans *= (n - end + 1);
    ans %= mod;
    cout << ans;
    return 0;
  }
  return 0;
}
