You've got a rectangular table with length a and width b and the infinite number of plates of radius r. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.

Input

A single line contains three space-separated integers a, b, r (1 ≤ a, b, r ≤ 100) — the table sides and the plates' radius, correspondingly.

Output

If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).

Examples

Input

5 5 2


Output

First


Input

6 7 4


Output

Second

Note

In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.

<image>

In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.

<image>
#include <bits/stdc++.h>
using namespace std;
const int bignumlen = 20200;
const int Blen = 8;
const long long base = 100000000;
struct bignum {
  int len;
  long long data[bignumlen];
  long long& operator[](int x) { return (data[x]); }
  const long long& operator[](int x) const { return (data[x]); }
  bignum() {
    memset(data, 0, sizeof(data));
    len = 0;
  }
  void clear() {
    for (int i = len; i >= 1; --i) data[i] = 0;
    len = 0;
  }
  int check(const bignum& a, const bignum& b) {
    if (a.len > b.len) return (0);
    if (b.len > a.len) return (1);
    for (int i = a.len; i >= 1; --i) {
      if (a.data[i] < b.data[i]) return (1);
      if (b.data[i] < a.data[i]) return (0);
    }
    return 2;
  }
  bool operator<(const bignum& b) { return (check(*this, b) == 1); }
  bool operator>(const bignum& b) { return (check(*this, b) == 0); }
  bool operator<=(const bignum& b) { return (check(*this, b) >= 1); }
  bool operator>=(const bignum& b) { return (check(*this, b) % 2 == 0); }
  bool operator!=(const bignum& b) { return (check(*this, b) != 2); }
  bool operator==(const bignum& b) { return (check(*this, b) == 2); }
  bignum operator=(const bignum& x) {
    for (int i = x.len + 1; i <= len; ++i) data[i] = 0;
    for (int i = 1; i <= x.len; ++i) data[i] = x.data[i];
    len = x.len;
    return *this;
  }
  bignum operator=(long long x) {
    for (int i = len; i >= 0; --i) data[i] = 0;
    len = 0;
    while (x) {
      data[++len] = x % base;
      x /= base;
    }
    return *this;
  }
  bignum(long long x) {
    memset(data, 0, sizeof(data));
    len = 0;
    (*this) = x;
  }
  bignum operator*(const bignum& b) {
    int i, j;
    bignum tmp;
    for (i = 1; i <= len; ++i)
      if (data[i] != 0)
        for (j = 1; j <= b.len; ++j)
          if (b.data[j] != 0) {
            tmp.data[i + j - 1] += data[i] * b.data[j];
            tmp.data[i + j] += tmp.data[i + j - 1] / base;
            tmp.data[i + j - 1] %= base;
          }
    tmp.len = len + b.len - 1;
    while (tmp.data[tmp.len + 1]) tmp.len++;
    return tmp;
  }
  bignum operator*(long long x) {
    int i;
    bignum tmp;
    for (i = 1; i <= len; ++i) tmp[i] = data[i] * x;
    tmp.len = len;
    for (i = 1; i <= len; ++i) {
      tmp[i + 1] += tmp[i] / base, tmp[i] %= base;
      if (tmp[i + 1] && i + 1 > tmp.len) tmp.len++;
    }
    return tmp;
  }
  bignum operator/(long long x) {
    int i;
    bignum tmp;
    long long y = 0;
    for (i = len; i >= 1; --i) {
      y = y * base + data[i];
      tmp[i] = y / x;
      y %= x;
    }
    tmp.len = len;
    while (tmp[tmp.len] == 0 && tmp.len > 1) tmp.len--;
    return tmp;
  }
  bignum operator/(const bignum& b) {
    if (b.len <= 1 && b[1] == 0) {
      printf("error! 被0除!");
      for (;;)
        ;
    }
    int i, l1 = (len - 1) * Blen, l2 = (b.len - 1) * Blen;
    long long x = data[len], y = b[b.len];
    while (x) x /= 10, l1++;
    while (y) y /= 10, l2++;
    bignum tmp, chu, B;
    chu = *this;
    B = b;
    for (i = 1; i * Blen <= l1 - l2; ++i) B *= base;
    for (i = 1; i <= (l1 - l2) % Blen; ++i) B *= 10;
    for (i = l1 - l2; i >= 0; --i) {
      x = 0;
      while (chu >= B) chu -= B, x++;
      tmp[i / Blen + 1] = tmp[i / Blen + 1] * 10 + x;
      B /= 10;
    }
    tmp.len = (l1 - l2) / Blen + 1;
    while (tmp.len >= 1 && !tmp[tmp.len]) tmp.len--;
    return tmp;
  }
  bignum operator+(const bignum& b) {
    bignum tmp;
    int i, l = max(len, b.len);
    for (i = 1; i <= l; ++i) tmp[i] = data[i] + b[i];
    for (i = 1; i <= l; ++i) tmp[i + 1] += tmp[i] / base, tmp[i] %= base;
    tmp.len = l;
    if (tmp[tmp.len + 1]) tmp.len++;
    return tmp;
  }
  bignum operator+(long long x) {
    bignum tmp;
    tmp = *this;
    tmp[1] += x;
    for (int i = 1; i <= len && tmp[i] >= base; ++i)
      tmp[i + 1] += tmp[i] / base, tmp[i] %= base;
    while (tmp[tmp.len + 1]) tmp.len++;
    return tmp;
  }
  bignum operator-(const bignum& b) {
    int i;
    bignum tmp;
    for (i = 1; i <= len; ++i) tmp.data[i] = data[i] - b.data[i];
    for (i = 1; i <= len; ++i) {
      if (tmp[i] < 0) tmp.data[i] += base, tmp.data[i + 1]--;
    }
    tmp.len = len;
    while (tmp[tmp.len] == 0 && tmp.len > 1) tmp.len--;
    return tmp;
  }
  bignum operator-(long long x) {
    bignum tmp;
    tmp = *this;
    tmp[1] -= x;
    for (int i = 1; i <= len && tmp[i] < 0; ++i) {
      tmp[i + 1] += (tmp[i] + 1) / base - 1;
      tmp[i] = (tmp[i] + 1) % base + base - 1;
    }
    while (!tmp[tmp.len] && tmp.len > 1) tmp.len--;
    return tmp;
  }
  long long operator%(long long x) {
    int i;
    long long y = 0;
    for (i = len; i >= 1; --i) y = (y * base + data[i]) % x;
    return y;
  }
  bignum operator%(const bignum& b) {
    if (b.len <= 1 && b[1] == 0) {
      printf("error! 被0 mod!");
      for (;;)
        ;
    }
    int i, l1 = (len - 1) * Blen, l2 = (b.len - 1) * Blen;
    long long x = data[len], y = b[b.len];
    while (x) x /= 10, l1++;
    while (y) y /= 10, l2++;
    bignum chu, B;
    chu = *this;
    B = b;
    for (i = 1; i * Blen <= l1 - l2; ++i) B *= base;
    for (i = 1; i <= (l1 - l2) % Blen; ++i) B *= 10;
    for (i = l1 - l2; i >= 0; --i) {
      while (chu >= B) chu -= B;
      B /= 10;
    }
    return chu;
  }
  bignum operator+=(const bignum& b) { return *this = (*this + b); }
  bignum operator*=(const bignum& b) { return *this = (*this * b); }
  bignum operator-=(const bignum& b) { return *this = (*this - b); }
  bignum operator/=(const bignum& b) { return *this = (*this / b); }
  bignum operator%=(const bignum& b) { return *this = (*this % b); }
  bignum operator*=(long long x) { return (*this = (*this * x)); }
  bignum operator+=(long long x) { return (*this = (*this + x)); }
  bignum operator-=(long long x) { return (*this = (*this - x)); }
  bignum operator/=(long long x) { return (*this = (*this / x)); }
  void read() {
    char c[bignumlen * Blen + 10];
    scanf("%s", c + 1);
    int l = strlen(c + 1);
    (*this).clear();
    long long x;
    for (int i = 1; i <= (l - 1) / Blen + 1; ++i) {
      x = 0;
      for (int j = l - Blen * i + 1; j <= l - Blen * i + Blen; ++j)
        if (j >= 1) x = x * 10 + c[j] - 48;
      data[++len] = x;
    }
  }
  void write() {
    printf("%lld", data[len]);
    for (int i = len - 1; i >= 1; --i) printf("%0*lld", Blen, data[i]);
  }
};
bignum gcd(const bignum& A, const bignum& B) {
  bignum a = A, b = B, res = 1;
  while (!(a[1] & 1) && !(b[1] & 1)) a /= 2, b /= 2, res *= 2;
  for (;;) {
    if (a.len == 1 && a[1] == 0) return b * res;
    if (b.len == 1 && b[1] == 0) return a * res;
    while (!(a[1] & 1)) a /= 2;
    while (!(b[1] & 1)) b /= 2;
    if (a > b)
      a -= b;
    else
      b -= a;
  }
}
int main() {
  string s;
  double num;
  long long a, b, r;
  while (cin >> a >> b >> r) {
    r *= 2;
    if (a >= r && b >= r)
      cout << "First\n" << endl;
    else
      cout << "Second\n" << endl;
  }
}
