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>
struct strgrp {
  strgrp(long long l, long long c) : lengthZero(l), cnt(c) {}
  long long lengthZero, cnt;
};
const long long MOD = 1e9 + 7;
std::string s;
std::vector<long long> blocks;
std::stack<strgrp> endZero;
int main() {
  std::ios::sync_with_stdio(false);
  std::cout.tie(nullptr);
  std::cin.tie(nullptr);
  std::cin >> s;
  int n = s.length();
  if (s.find('1') == std::string::npos) {
    std::cout << n << '\n';
    return 0;
  }
  for (int iChr = 0; iChr < n; iChr++) {
    if (s[iChr] == '1')
      blocks.emplace_back(0);
    else {
      int jChr = iChr;
      while (jChr < n && s[jChr] == '0') jChr++;
      blocks.emplace_back(jChr - iChr);
      iChr = jChr - 1;
    }
  }
  int mult = 1;
  if (blocks[0] != 0) {
    mult = (blocks[0] + 1);
    blocks.erase(blocks.begin());
  }
  long long maxOne = 1, maxZero = 0;
  long long gen = 0;
  for (long long block : blocks) {
    if (block == 0) {
      maxOne = (maxOne + maxZero) % MOD;
      maxZero = 0;
      gen = (gen + maxOne) % MOD;
      endZero.emplace(0, maxOne);
    }
    long long nxtMax = 0;
    while (!endZero.empty() && endZero.top().lengthZero <= block) {
      nxtMax = (nxtMax + endZero.top().cnt) % MOD;
      maxZero =
          (maxZero + (block - endZero.top().lengthZero) * endZero.top().cnt) %
          MOD;
      endZero.pop();
    }
    endZero.emplace(block, nxtMax);
  }
  while (!endZero.empty()) {
    gen = (gen + endZero.top().cnt * blocks.back()) % MOD;
    endZero.pop();
  }
  std::cout << (gen * mult) % MOD << '\n';
  return 0;
}