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>
const int MAX_N = 1000000;
const int MOD = 1000000007;
std::vector<int> groups;
int table[1 + MAX_N], dp[1 + MAX_N];
std::deque<int> tablePositions;
int main() {
  std::string s;
  std::cin >> s;
  groups.push_back(0);
  int cnt = 0;
  for (int i = 0; i < s.size(); ++i)
    if (s[i] == '0')
      ++cnt;
    else {
      groups.push_back(cnt);
      cnt = 0;
    }
  groups.push_back(cnt);
  dp[1] = groups[1] + 1;
  for (int i = 2; i < groups.size(); ++i) {
    if (i < groups.size() - 1) dp[i] = dp[i - 1];
    for (int lastDigit = 0; lastDigit <= groups[i]; ++lastDigit) {
      dp[i] = (dp[i] + dp[i - 1]) % MOD;
      while (!tablePositions.empty() && tablePositions.front() < lastDigit)
        tablePositions.pop_front();
      if (i < groups.size() - 1 && !tablePositions.empty())
        dp[i] = (dp[i] + MOD - table[tablePositions.front()]) % MOD;
    }
    tablePositions.push_front(groups[i]);
    table[groups[i]] = dp[i - 1];
  }
  if (groups.size() == 2) dp[groups.size() - 1]--;
  printf("%d", dp[groups.size() - 1]);
  return 0;
}