#include <iostream>
#include <cstring>
#include <deque>
using namespace std; #define SIZE 230
#define BACK 1
#define AWAY 0 int DP[SIZE][SIZE][2];
bool visits[SIZE];
int vals[SIZE];
deque< int > tree[SIZE]; int num, steps; void dfs( int u ){ visits[u] = true;
const int len = tree[u].size(); for( int i = 0; i <= steps; ++i )
DP[u][i][BACK] = DP[u][i][AWAY] = vals[u]; for( int i = 0; i < len; ++i ){ int son = tree[u][i]; if( visits[son] )
continue; dfs( son ); for( int s = steps; s >= 0; --s ){
for( int ss = 0; ss <= s; ++ss ){
/*
从 u 出发,回到 u。须要多走两步 u->son,son->u,
分配给 son 子树 ss 步,其它子树 s - ss 步。都返回.
*/
DP[u][s + 2][BACK] = max( DP[u][s + 2][BACK],
DP[u][s - ss][BACK] + DP[son][ss][BACK] ); /*
不回 u (去 u 的其它子树)。在 son 返回.
*/
DP[u][s + 2][AWAY] = max( DP[u][s + 2][AWAY],
DP[u][s - ss][AWAY] + DP[son][ss][BACK] ); /*
先遍历 u 的其它子树,回到 u 后,遍历 son 子树,
在当前子树 son 不返回,多走一步.
*/
DP[u][s + 1][AWAY] = max( DP[u][s + 1][AWAY],
DP[u][s - ss][BACK] + DP[son][ss][AWAY] ); }
}
}
} int main(){ int u, v; while( cin >> num >> steps ){ memset( DP, 0, sizeof( DP ) );
memset( visits, false, sizeof( visits ) ); for( int i = 1; i <= num; ++i )
tree[i].clear(); for( int i = 1; i <= num; ++i )
cin >> vals[i]; for( int i = 1; i <= num - 1; ++i ){ cin >> u >> v; tree[u].push_back( v );
tree[v].push_back( u ); } dfs( 1 ); cout << max( DP[1][steps][BACK], DP[1][steps][AWAY] ) << endl; } return 0; }