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 long long MOD = 1000000007;
signed main() {
  std::string s;
  std::cin >> s;
  long long n = s.length();
  long long cur = 0;
  std::vector<long long> blocks;
  for (long long i = 0; i <= n; i++) {
    if (i == n || s[i] == '1') {
      blocks.push_back(cur);
      cur = 0;
    }
    cur += s[i] == '0';
  }
  long long m = blocks.size();
  if (m == 1) {
    std::cout << n << std::endl;
    return 0;
  }
  long long mult = ((blocks.front() + 1) * (blocks.back() + 1)) % MOD;
  blocks.pop_back();
  blocks.push_back(0);
  long long dp[m + 1];
  long long extra[m + 1];
  long long firstg[m + 1];
  dp[m] = extra[m] = 0;
  dp[m - 1] = 1;
  extra[m - 1] = 0;
  firstg[m - 1] = -1;
  for (long long i = m - 2; i >= 0; i--) {
    firstg[i] = i + 1;
    while (firstg[i] != -1 && blocks[firstg[i]] <= blocks[i])
      firstg[i] = firstg[firstg[i]];
  }
  for (long long i = m - 2; i >= 1; i--) {
    dp[i] = 1 + (blocks[i] + 1) * dp[i + 1];
    dp[i] %= MOD;
    long long j = firstg[i];
    extra[i] = 0;
    if (j != -1) {
      extra[i] = (blocks[j] - blocks[i]) * dp[j + 1] + extra[j];
      extra[i] %= MOD;
    }
    dp[i] = (dp[i] + extra[i]) % MOD;
  }
  std::cout << (mult * dp[1]) % MOD << std::endl;
  return 0;
}