CodeForces 811A ——Vladik and Courtesy——暴力，模拟

A. Vladik and Courtesy

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

At regular competition Vladik and Valera won a and b candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.

More formally, the guys take turns giving each other one candy more than they received in the previous turn.

This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.

Input

Single line of input data contains two space-separated integers a, b (1 ≤ a, b ≤ 109) — number of Vladik and Valera candies respectively.

Output

Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.

Examples

input

1 1

output

Valera

input

7 6

output

Vladik

Note

Illustration for first test case:

Illustration for second test case:

两个人分别有a，b颗糖，第一个人先少一颗，然后第二个人少两颗，然后再第一个人少三颗…………以此类推。。看看谁的糖不够拿了

丢脸的一题。。。。看样例脑补题意失败。。。读错题了。。。刚开始用二分。。。其实直接暴力模拟或者算等差数列就好了。。

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<vector>#include<set>#include<map>#include<queue>#include<stack>#define LL long longusing namespace std;const int N = 100010;int main(){    LL a, b;    while(scanf("%lld%lld", &a, &b) == 2){        for(LL i = 1; a >= 0 && b >= 0; i ++){            if(i % 2){                a -= i;            }            else{                b -= i;            }        }         if (a < 0)             printf("Vladik\n");         else             printf("Valera\n");    }    return 0;}