Fancy Sequence LeetCode solution in C++

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int mod97 = 1000000007;
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/**
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Calculates multiplicative inverse
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*/
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unsigned long modPow(unsigned long x, int y) {
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  unsigned long tot = 1, p = x;
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  for (; y; y >>= 1) {
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    if (y & 1) tot = (tot * p) % mod97;
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    p = (p * p) % mod97;
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  }
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  return tot;
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}
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class Fancy {
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 public:
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  unsigned long seq[100001];
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  unsigned int length = 0;
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  unsigned long increment = 0;
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  unsigned long mult = 1;
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  Fancy() {
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    ios_base::sync_with_stdio(false);
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    cin.tie(NULL);
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  }
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  void append(int val) {
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    seq[length++] = (((mod97 + val - increment) % mod97) * modPow(mult, mod97 - 2)) % mod97;
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  }
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  void addAll(int inc) {
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    increment = (increment + inc % mod97) % mod97;
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  }
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  void multAll(int m) {
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    mult = (mult * m % mod97) % mod97;
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    increment = (increment * m % mod97) % mod97;
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  }
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  int getIndex(int idx) {
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    if (idx >= length) {
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      return -1;
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    } else {
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      return ((seq[idx] * mult) % mod97 + increment) % mod97;
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    }
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  }
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};