Yig'indi chegarani kesib o'tganda

Natural son $n$ berilgan. $1 + 2 + \ldots + k \ge n$ bo'ladigan eng kichik natural $k$ ni toping. $k$ ni va shu paytdagi yig'indini chiqaring.

Bitta butun son $n$ ($1 \le n \le 10^{12}$).

Bitta qatorda ikkita butun son: $k$ va $1 + 2 + \ldots + k$.

Nechta qadam kerakligi oldindan noma'lum — while sikli: yig'indi $n$ dan kichik ekan, navbatdagi sonni qo'shing. Birinchi misolda: $1 + 2 + 3 + 4 = 10 \ge 10$, javob 4 10; ikkinchisida $k = 5$, yig'indi $15$.

Yig'indi $10^{12}$ dan oshadi — long long / int64 kerak.