B"LuoTianyi gives you a tree with values in its vertices, and the root of the tree is vertex 1 . In one operation, you can change the value in one vertex to any non-negative integer. Now you need to find the minimum number of operations you need to perform to make each path from the root to leaf ^{ dagger} has a bitwise XOR value of zero. ^{ dagger} A leaf in a rooted tree is a vertex that has exactly one neighbor and is not a root. The first line contains a single integer n ( 2 <= n <= 10^5 ) -- the number of vertices in the tree. The second line contains n integers a_1, a_2, ldots, a_n ( 1 <= a_i <= 10^9 ), the i -th number represents the value in the i -th vertex. Next n xe2 x88 x921 lines describe the edges of the tree. The i -th line contains two integers u_i and v_i ( 1 <= u_i,v_i <= n, u_i neq v_i ) -- the vertices connected by an edge of the tree. It's guaranteed that the given edges form a tree. Print a single integer -- the minimum number of operations. The tree in the first example: If we change the value in the vertex 2 to 3 , the value in the vertex 5 to 4 , and the value in the vertex 6 to 6 , then the tree will be ok. The bitwise XOR from the root to the leaf 2 will be 3 oplus 3=0 . The bitwise XOR from the root to the leaf 5 will be 4 oplus 7 oplus 3=0 . The bitwise XOR from the root to the leaf 6 will be 6 oplus 5 oplus 3=0 . The tree in the second example: If we change the value in the vertex 2 to 4 , the value in the vertex 3 to 27 , and the value in the vertex 6 to 20 , then the tree will be ok. The bitwise XOR from the root to the leaf 6 will be 20 oplus 19 oplus 7=0 . The bitwise XOR from the root to the leaf 8 will be 11 oplus 27 oplus 4 oplus 19 oplus 7=0 . The bitwise XOR from the root to the leaf 4 will be 16 oplus"...