CF633A.Ebony and Ivory

Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.

For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.

但丁正在与“救世主”战斗。在能用剑与其交战之前，他必须先击破其护盾。他有两把枪：黑檀木（Ebony）和象牙（Ivory），每把枪均可发射任意非负整数发子弹。

每一发命中护盾的子弹中，黑檀木造成 $a$ 点伤害，象牙造成 $b$ 点伤害。为了击破护盾，但丁必须恰好造成 $c$ 点伤害。请判断这是否可行。

The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.

输入的第一行包含三个整数 $a$、$b$、$c$（$1 ≤ a, b ≤ 100$，$1 ≤ c ≤ 10\,000$），分别表示 Ebony 枪和 Ivory 枪每次造成的伤害量，以及击破护盾所需的总伤害量。

Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.

如果但丁能对护盾造成恰好 c 点伤害，则输出 "Yes"（不带引号）；否则输出 "No"（不带引号）。

In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.

在第二个样例中，但丁可以使用“黑檀木”发射 1 发子弹、“象牙”发射 2 发子弹，恰好造成 $1\cdot3 + 2\cdot2 = 7$ 点伤害。在第三个样例中，但丁可以使用“黑檀木”发射 1 发子弹、“象牙”不发射子弹，造成 $1\cdot6 + 0\cdot11 = 6$ 点伤害。