一道尺取法 （连续最值）

#include <iostream>#include <algorithm>using namespace std;int main(){int a[365];//now first build an arrayint n;cin >> n;    for (int i = 1;i <= n;i ++)    cin >> a[i]; int i = 1,j = 1; int sum = 0; int minTime = 1000; while (j != n+1){ sum += a[j++]; minTime = min(abs(360-sum*2),minTime); while (sum >= 180){ sum -= a[i++]; minTime = min(abs(360-sum*2),minTime); } } cout << minTime << endl;return 0;}

A. Pizza Separation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut inton pieces. The i-th piece is a sector of angle equal to a_i. Vasya and Petya want to divide all pieces of pizza into twocontinuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.

Input

The first line contains one integer n (1 ≤ n ≤ 360)  — the number of pieces into which the delivered pizza was cut.

The second line contains n integers a_i (1 ≤ a_i ≤ 360)  — the angles of the sectors into which the pizza was cut. The sum of all a_i is 360.

Output

Print one integer  — the minimal difference between angles of sectors that will go to Vasya and Petya.

Examples

Input

4
90 90 90 90

Output

0

Input

3
100 100 160

Output

40

Input

1
360

Output

360

Input

4
170 30 150 10

Output

0