#include <stdio.h>

 #include <algorithm>

 #include <string.h>

 using namespace std;

 const int N =  + ;

 typedef long long ll;

 const int X = ;

 const int mod = (int)1e9 + ;

 char s[N];

 int m,len,pw[N];

 int H[N],pos;

 struct node

 {

     int id,hash;

     bool operator < (const node & temp) const

     {

         return hash == temp.hash ? id < temp.id : hash < temp.hash;

     }

 }p[N];

 bool solve(int L)

 {

     int cnt = ;

     pos = -;

     for(int i=;i+L-<=len;i++)

     {

         int id = i;

         int hash = ((ll)H[i] - (ll)H[i+L]*pw[L]) % mod;

         if(hash < ) hash += mod; // 注意这里！

         p[i] = (node){id, hash};

     }

     sort(p+, p++ len - L + );

     for(int i=;i<=len-L+;i++)

     {

         if(i ==  || p[i].hash != p[i-].hash) cnt = ;

         if(++cnt >= m) pos = max(pos, p[i].id);

     }

     return pos != -;

 }

 int main()

 {

     pw[] = ;

     for(int i=;i<N;i++) pw[i] = 1LL*pw[i-] * X % mod;

     while(scanf("%d",&m) ==  && m)

     {

         scanf("%s",s+);

         len = strlen(s+);

         for(int i=len;i>=;i--) H[i] = (1LL*H[i+] * X + s[i] - 'a') % mod;

         int l = , r = len, ans = -;

         while(l <= r)

         {

             int mid = l + r >> ;

             if(solve(mid)) ans = mid, l = mid + ;

             else r = mid - ;

         }

         solve(ans);

         if(ans != -) printf("%d %d\n",ans, pos-);

         else puts("none");

     }

     return ;

 }

 #include <stdio.h>

 #include <algorithm>

 #include <string.h>

 using namespace std;

 const int N =  + ;

 typedef long long ll;

 /**

  *    sa[i]:表示排在第i位的后缀的起始下标

  *    rank[i]:表示后缀suffix(i)排在第几

  *    height[i]:sa[i-1] 与 sa[i]的LCP(最长公共前缀)值

  *

  * */

  /*

     如果整数的话模板改成int.

     加一个数a[n] = 0 。 这样他的排名是第一个。

     construct(a,n+1);

     字符串的话。

     len = strlen(str);

     construct(s,strlen(s)+1);

     排名第0的是个空字符串。

     height[i]:sa[i-1] 与 sa[i]的LCP(最长公共前缀)值

     所以height[1] = 0;

     rank[len] = 0;

     sa[0] = len;

  */

 int sa[N],rnk[N],height[N];

 void construct(const char *s,int n,int m = ) {

     static int t1[N],t2[N],c[N];

     int *x = t1,*y = t2;

     int i,j,k,p,l;

     for (i = ; i < m; ++ i) c[i] = ;

     for (i = ; i < n; ++ i) c[x[i] = s[i]] ++;

     for (i = ; i < m; ++ i) c[i] += c[i - ];

     for (i = n - ; i >= ; -- i) sa[--c[x[i]]] = i;

     for (k = ; k <= n; k <<= ) {

         p = ;

         for (i = n - k; i < n; ++ i) y[p++] = i;

         for (i = ; i < n; ++ i) if (sa[i] >= k) y[p++] = sa[i] - k;

         for (i = ; i < m; ++ i) c[i] = ;

         for (i = ; i < n; ++ i) c[x[y[i]]] ++;

         for (i = ; i < m; ++ i) c[i] += c[i - ];

         for (i = n - ; i >= ; -- i) sa[--c[x[y[i]]]] = y[i];

         std::swap(x,y);

         p = ; x[sa[]] = ;

         for (i = ; i < n; ++ i)

             x[sa[i]] = y[sa[i - ]] == y[sa[i]]

                 && y[sa[i - ] + k] == y[sa[i] + k] ? p - : p ++;

         if (p >= n) break;

         m = p;

     }

     for (i = ; i < n; ++ i) rnk[sa[i]] = i;

     for (i = ,l = ; i < n; ++ i) {

         if (rnk[i]) {

             j = sa[rnk[i] - ];

             while (s[i + l] == s[j + l]) l++;

             height[rnk[i]] = l;

             if (l) l--;

         }

     }

 }

 int m,pos,len;

 char s[N];

 bool solve(int L)

 {

     // 用后缀数组的话，如果重复长度就是最长的长度，需要特判

     if(m ==  && L == len) {pos = ; return ;}

     pos = -;

     int cnt = ;

     int temp = -; // 要用一个temp来记录当前cnt>=m的最大的位置

     for(int i=;i<=len;i++)

     {

         if(height[i] >= L) cnt++, temp = max(temp, sa[i]);

         else cnt = , temp = sa[i];

         if(cnt >= m) pos = max(pos, temp);

     }

     return pos != -;

 }

 int main()

 {

     while(scanf("%d",&m) ==  && m)

     {

         scanf("%s",s);

         len = strlen(s);

         construct(s, len+);

         int l = , r = len;

         int ans = -;

         while(l <= r)

         {

             int mid = l + r >> ;

             if(solve(mid)) ans = mid, l = mid + ;

             else r = mid - ;

         }

         solve(ans);

         if(ans != -) printf("%d %d\n",r,pos);

         else puts("none");

     }

     return ;

 }