/*

 * @Author: mjt

 * @Date:   2018-10-17 20:55:42

 * @Last Modified by:   mjt

 * @Last Modified time: 2018-10-17 21:37:10

 */

 #include<cstdio>

 #include<algorithm>

 #include<cstring>

 #include<cmath>

 #include<iostream>

 #include<cctype>

 #include<set>

 #include<vector>

 #include<queue>

 #include<map>

 #define fi(s) freopen(s,"r",stdin);

 #define fo(s) freopen(s,"w",stdout);

 using namespace std;

 typedef long long LL;

 inline int read() {

     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;

     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;

 }

 class CountryGroupHard{

 public:

     int f[];

     string solve(vector <int> a) {

         int n = a.size();

         f[n] = ;

         for (int i = n - ; i >= ; --i) {

             if (a[i] == ) {

                 int ok = -;

                 for (int j = ; (i + j) <= n; ++j) {

                     if (a[i + j - ] != ) { // 最近的有数的位置

                         if (ok != - && ok != a[i + j - ]) break;

                         else ok = a[i + j - ];

                     }

                     if (ok == - || ok == j) f[i] += f[i + j];

                 }

             }

             else {

                 bool ok = (i + a[i] > n); // 要求后面的a[i]个相等

                 for (int j = i; j < n && j < i + a[i]; ++j)

                     if (a[j] && a[j] != a[i]) ok = true;

                 if (!ok) f[i] = f[i + a[i]];

             }

             if (f[i] >= ) f[i] = ;

         }

         return f[] >  ? "Insufficient" : "Sufficient";

     }

 };