A. Lever

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time limit per test

memory limit per test

256>

input

standard>

output

standard>

You have a description of a lever as string s. We'll represent the string length as record |s|, then the lever looks as a horizontal bar with weights of length |s| - 1 with exactly one pivot. We will assume that the bar is a segment on the Ox axis between points 0 and |s| - 1.

The decoding of the lever description is given below.

If the i-th>^", that means that at coordinate i there is the pivot under the bar.
If the i-th>=", that means that at coordinate i there is nothing lying on the bar.
If the i-th character of the string equals digit c (1-9), that means that at coordinate i there is a weight of mass c on the bar.

Your task is, given the lever description, print if it will be in balance or not. Assume that the bar doesn't weight anything. Assume that the bar initially is in balance then all weights are simultaneously put on it. After that the bar either tilts to the left, or tilts to the right, or is in balance.

Input

The first line contains the lever description as a non-empty string s (3 ≤ |s| ≤ 106), consisting of digits (1-9) and characters "^" and "=". It is guaranteed that the line contains exactly one character "^". It is guaranteed that the pivot of the lever isn't located in any end of the lever bar.

To solve the problem you may need 64-bit integer numbers. Please, do not forget to use them in your programs.

Output

Print "left" if the given lever tilts to the left, "right" if it tilts to the right and "balance",>

Sample test(s)

input

=^==

output

balance

input

9===^==1

output

left

input

2==^7==

output

right

input

41^52==

output

balance

Note

As you solve the problem, you may find the following link useful to better understand how a lever functions:http://en.wikipedia.org/wiki/Lever.

The pictures to the examples:

解题说明：此题为一道物理天平题，题目中标记为^的为支撑架所在的位置，=代表天平的杆，左右的数字代表天平上每个砝码的重量，学过物理的都知道天平两端力矩乘上砝码质量加和相等天平就平衡，否则按照加和大的那一侧倾斜。反应到程序中，就判断加和即可。

#include<iostream>#include<cstdio>#include<cmath>using namespace std;int main(){char s[1000008];int i, j, L1 = 1, L2 = 1;long long a1 = 0, a2 = 0;scanf("%s", s);for (i = 0; s[i] != '\n'; i++){if (s[i] == '^'){break;}}j = i;for (i = j - 1; i >= 0; i--)  {if (s[i] != '='){a1 += (s[i] - 48)*L1;}L1++;}for (i = j + 1; s[i] != '\0'; i++)  {if (s[i] != '='){a2 += (s[i] - 48)*L2;}L2++;}if (a1 > a2){printf("left\n");}else if (a1 < a2){printf("right\n");}else{printf("balance\n");}return 0;}

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