/*********************************************************
*            ------------------                          *
*   author AbyssalFish                                   *
**********************************************************/
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<numeric>
using namespace std;/*
容斥原理去重复
左边是枚举出来的,保证不等
可能出现重复的情况有：
d 和 c同时出现
case 1 d+c == d-c -> c == 0其中一个出现
前提
a or b != c or dcase2  a+d == d-c, -> a == -c , 必要条件 a != c -> c != 0 如果不满足这个条件：找到的可能是d == d
case3  d+b == d-c, -> b == -c
case 2 3互斥case4  a+c == d-c, -> a == d-2*c, 必要条件 a != d -> c != 0
case5  c+b == d-c -> b == d-2*c
case 4 5互斥case (2,3,4,5) 1互斥
*/const int maxn = 1e3, INF = 536870911+5;
int n;
int e[maxn+1], tab[maxn*maxn>>1];
bool vis[maxn];int solve()
{sort(e,e+n);int i, j, a, c, d, k;int sz = 0;for(i = 1; i < n; i++){a = e[i];for(j = 0; j < i; j++){tab[sz++] = a+e[j];}}sort(tab,tab+sz);e[n] = e[n]-1;for(j = 0; j < n; j++) {c = -e[j];vis[j] = (*lower_bound(e,e+n,c) == c); // c == 0 || -c == a or b
    }tab[sz+1] = tab[sz] = tab[sz-1]-1;for(i = n; i--; ){d = e[i];for(j = 2; j < n; j++)if(i!=j){c = e[j];a = d-c;k = lower_bound(tab,tab+sz,a)-tab;if(c){if(vis[j]) k++;if(*lower_bound(e,e+n,a-c) == a-c) k++;}else k++;if(a == tab[k]) return d;}}return INF;
}//#define LOCAL
int main()
{
#ifdef LOCALfreopen("in.txt","r",stdin);
#endifwhile(scanf("%d",&n),n){for(int i = 0; i < n; i++) scanf("%d", e+i);int res = solve();if(res < INF) printf("%d\n", res);else puts("no solution");}return 0;
}

int solve()
{sort(e,e+n);int i, j, sum, lb, ub, tsum;for(i = n; i--; ){for(j = 2; j < n; j++)if(i!=j){sum = e[i]-e[j];for(lb = 0,ub = j-1; lb < ub; ){ //状态是DAGtsum = e[lb]+e[ub];if(tsum == sum) return e[i];tsum > sum ? ub--:lb++; //O(n)
            }}}return INF;
}