CodeForces

B. Painting Eggs

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Bitlandians are quite weird people. They have very peculiar customs.

As is customary, Uncle J. wants to have n eggs painted for Bitruz (an ancient Bitland festival). He has asked G. and A. to do the work.

The kids are excited because just as is customary, they're going to be paid for the job!

Overall uncle J. has got n eggs. G. named his price for painting each egg. Similarly, A. named his price for painting each egg. It turns out that for each egg the sum of the money both A. and G. want for the painting equals 1000.

Uncle J. wants to distribute the eggs between the children so as to give each egg to exactly one child. Also, Uncle J. wants the total money paid to A. to be different from the total money paid to G. by no more than500.

Help Uncle J. Find the required distribution of eggs or otherwise say that distributing the eggs in the required manner is impossible.

Input

The first line contains integer n (1 ≤ n ≤ 10⁶) — the number of eggs.

Next n lines contain two integers a_i and g_i each (0 ≤ a_i, g_i ≤ 1000; a_i + g_i = 1000):a_i is the price said by A. for thei-th egg and g_i is the price said by G. for thei-th egg.

Output

If it is impossible to assign the painting, print "-1" (without quotes).

Otherwise print a string, consisting of n letters "G" and "A". Thei-th letter of this string should represent the child who will get thei-th egg in the required distribution. Letter "A" represents A. and letter "G" represents G. If we denote the money Uncle J. must pay A. for the painting as S_a, and the money Uncle J. must pay G. for the painting asS_g, then this inequality must hold:|S_a  -  S_g|  ≤  500.

If there are several solutions, you are allowed to print any of them.

Examples

Input

21 999999 1

Output

AG

Input

3400 600400 600400 600

Output

AGA

题目大意：就是有两个序列，A和G，A[i]+G[i]为1000，每次从序列中一个取出一个元素，同时从A序列取的元素代数和sum_a和从G序列取的元素代数和sum_g之差不超过500，如果存在打印取的顺序，没有的话就输出-1

大体想法：贪心算法，既然使取元素两边之差不超过500，则可以使每一步取完之后的之差不超过500，这样显然是满足条件的。

#include <iostream>#include <cstdio>#include <algorithm>#include <string>using namespace std;int main() {    int n;        scanf("%d",&n);        int a = 0;    int g = 0;    bool flag = true;    string s = "";    int index = 0;        for(int i = 0;i < n;i++) {        int x,y;                scanf("%d %d",&x,&y);        if(a+x-g <= 500) {            a += x;            s.append("A");        } else if(g+y-a <= 500){            g += y;            s.append("G");        } else {            flag = false;            break;        }    }    if(flag) {        cout << s << "\n";    } else {        printf("-1\n");    }        return 0;}