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 M_MAX = 1000002;
const int MOD = 1000000007;
int n;
string s;
int m;
int a[M_MAX];
int dp[M_MAX];
int st[M_MAX];
int lg;
int val;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  s = " " + s;
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '1') {
      a[++m] = sum;
      sum = 0;
    } else
      sum++;
  }
  a[++m] = sum;
  if (m == 1) {
    cout << n << "\n";
    return 0;
  }
  dp[0] = 1;
  for (int i = 1; i <= m; i++) {
    if (i > 1) {
      while (lg > 0 && a[st[lg]] <= a[i]) {
        val -= 1LL * (a[st[lg]] + 1) * dp[st[lg] - 1] % MOD;
        if (val < 0) val += MOD;
        if (lg > 1) {
          val += 1LL * (a[st[lg]] + 1) * dp[st[lg - 1] - 1] % MOD;
          if (val >= MOD) val -= MOD;
        }
        lg--;
      }
      if (lg > 0) {
        val -= 1LL * (a[i] + 1) * dp[st[lg] - 1] % MOD;
        if (val < 0) val += MOD;
      }
      val += 1LL * (a[i] + 1) * dp[i - 1] % MOD;
      if (val >= MOD) val -= MOD;
      st[++lg] = i;
    }
    dp[i] = val;
    dp[i] += 1LL * dp[0] * (a[1] + 1) % MOD;
    if (dp[i] >= MOD) dp[i] -= MOD;
  }
  cout << 1LL * dp[m - 1] * (a[m] + 1) % MOD << "\n";
  return 0;
}
