Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.

Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).

What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?

Single line of the input contains three integers x, y and m ( - 10¹⁸ ≤ x, y, m ≤ 10¹⁸).

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64dspecifier.

Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.

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#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <stack>

#include <vector>

#include <queue>

#include <cstring>

#include <string>

#include <map>

#include <set>

using namespace std;

#define LL long long

const int BufferSize = 1 << 16;

char buffer[BufferSize], *Head, *Tail;

inline char Getchar() {

    if(Head == Tail) {

        int l = fread(buffer, 1, BufferSize, stdin);

        Tail = (Head = buffer) + l;

    }

    return *Head++;

}

LL read() {

    LL x = 0, f = 1; char c = getchar();

    while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }

    while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }

    return x * f;

}

LL x, y, m;

int main() {

	x = read(); y = read(); m = read();

	if(x > y) swap(x, y);

	if(x <= 0 && y <= 0 && y < m) return puts("-1"), 0;

	if(y >= m) return puts("0"), 0;

	LL cnt = abs(x) % y ? abs(x) / y + 1 : abs(x) / y;

	x += cnt * y; if(x > y) swap(x, y);

	while(y < m) {

		x = x + y;

		if(x > y) swap(x, y);

		cnt++;

	}

	printf("%I64d\n", cnt);

	return 0;

}