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long long fun(long long a) { // sum of all natural from 1 to a
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  long long b = a * (a + 1) / 2;
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  return b;
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}
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class Solution {
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 public:
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  int reachNumber(int target) {
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    long long i = 1, j = pow(10, 5), x = abs(target),
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              ans = 0; // for -ve or +ve positive of a number the minimum no. of
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                       // steps from the origin will be same
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    while (i <= j) { // binary search to search if x is continous sum of some
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                     // natural number starting from 1
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      long long m = (i + j) / 2;
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      if (fun(m) == x) {
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        ans = m;
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      }
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      if (x > fun(m)) {
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        i = m + 1;
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      } else {
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        j = m - 1;
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      }
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    }
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    if (ans != 0) { // If we found our ans return it
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      return ans;
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    } else {
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      // in this for loop i have set the limit too high it can be less then 10^6
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      // as max value j can be is 44723, so loop will never run fully, whatever
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      // the value of j will be, loop will maximum run for 3-10 iterations
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      for (int l = j + 1; l < 100000; l++) { // in the end of binary search we get the value of j(or
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                                             // high end) as the position of the number(in the
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                                             // sequence of continous sum of natural number from 1,
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                                             // i.e. 1, 3,6,10........) whose value is just less
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                                             // than x(searching element)
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        if ((fun(l) - x) % 2 == 0) { // as the total step will be more than x if we go backward from
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                                     // zero, thing to note is that if we go -ve direction we also
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                                     // have to come back so we covering even distance
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          ans = l;                   // when the first fun(l) - x is even that l is our minimum jump
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          break; // no need to search further
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        }
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      }
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    }
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    return ans;
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  }
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};