Digital collectible card games have become very popular recently. So Vova decided to try one of these.

Vova has n cards in his collection. Each of these cards is characterised by its power pi, magic number ci and level li. Vova wants to build a deck with total power not less than k, but magic numbers may not allow him to do so — Vova can't place two cards in a deck if the sum of the magic numbers written on these cards is a prime number. Also Vova cannot use a card if its level is greater than the level of Vova's character.

At the moment Vova's character's level is 1. Help Vova to determine the minimum level he needs to reach in order to build a deck with the required total power.

Input

The first line contains two integers $n$ and $k$ $(1 ≤ n ≤ 100, 1 ≤ k ≤ 100000)$.

Then n lines follow, each of these lines contains three numbers that represent the corresponding card: $p_i$, $c_i$ and $l_i$ $(1 ≤ p_i ≤ 1000, 1 ≤ c_i ≤ 100000, 1 ≤ l_i ≤ n)$.

Output

If Vova won't be able to build a deck with required power, print $ - 1$. Otherwise print the minimum level Vova has to reach in order to build a deck.

 #include<bits/stdc++.h>

 const int maxn = ;

 const int maxm = ;

 const int maxNum = ;

 const int INF = 2e9;

 struct Edge

 {

     int u,v,f,c;

     Edge(int a=, int b=, int c=, int d=):u(a),v(b),f(c),c(d) {}

 }edges[maxm];

 bool pr[maxNum];

 int n,K,ans,L,R,mid,tag,cnt,S,T;

 int p[maxn],c[maxn],l[maxn],sv[maxn];

 int edgeTot,head[maxn],nxt[maxm],lv[maxn];

 int read()

 {

     char ch = getchar();

     int num = , fl = ;

     for (; !isdigit(ch); ch = getchar())

         if (ch=='-') fl = -;

     for (; isdigit(ch); ch = getchar())

         num = (num<<)+(num<<)+ch-;

     return num*fl;

 }

 void addedge(int u, int v, int c)

 {

     edges[edgeTot] = Edge(u, v, , c), nxt[edgeTot] = head[u], head[u] = edgeTot++;

     edges[edgeTot] = Edge(v, u, , ), nxt[edgeTot] = head[v], head[v] = edgeTot++;

 }

 bool buildLevel()

 {

     memset(lv, , sizeof lv);

     std::queue<int> q;

     lv[S] = , q.push(S);

     for (int tmp; q.size(); )

     {

         tmp = q.front(), q.pop();

         for (int i=head[tmp]; i!=-; i=nxt[i])

         {

             int v = edges[i].v;

             if (!lv[v]&&edges[i].f < edges[i].c){

                 lv[v] = lv[tmp]+, q.push(v);

                 if (v==T) return true;

             }

         }

     }

     return false;

 }

 int fndPath(int x, int lim)

 {

     if (x==T) return lim;

     for (int i=head[x]; i!=-; i=nxt[i])

     {

         int v = edges[i].v, val;

         if (lv[x]+==lv[v]&&edges[i].f < edges[i].c){

             if ((val = fndPath(v, std::min(lim, edges[i].c-edges[i].f)))){

                 edges[i].f += val, edges[i^].f -= val;

                 return val;

             }else lv[v] = -;

         }

     }

     return ;

 }

 int dinic()

 {

     int ret = , val;

     while (buildLevel())

         while ((val = fndPath(S, INF)))

             ret += val;

     return ret;

 }

 int main()

 {

     n = read(), K = read();

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

         if (!pr[i]) for (int j=i+i; j<maxNum; j+=i)

             pr[j] = true;

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

         p[i] = read(), c[i] = read(), l[i] = read();

     ans = -, L = , R = , S = , T = n+;

     for (mid=(L+R)>>; L<=R; mid=(L+R)>>)

     {

         memset(head, -, sizeof head);

         sv[] = tag = cnt = edgeTot = ;

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

             if ((c[i]==&&p[i] > p[tag])&&(l[i] <= mid)) tag = i;

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

             if ((c[i]!=||i==tag)&&(l[i] <= mid)) sv[++sv[]] = i;

         for (int i=; i<=sv[]; cnt += p[sv[i]], i++)

             if (c[sv[i]]&) addedge(S, i, p[sv[i]]);

             else addedge(i, T, p[sv[i]]);

         for (int i=; i<=sv[]; i++)

             if (c[sv[i]]&) for (int j=; j<=sv[]; j++)

                 if ((~c[sv[j]]&)&&(!pr[c[sv[i]]+c[sv[j]]]))

                     addedge(i, j, INF);

         cnt -= dinic();

         if (cnt >= K) ans = mid, R = mid-;

         else L = mid+;

     }

     printf("%d\n",ans);

     return ;

 }