/** 
   * 第一步判断，遍历A树查找是否有等于B树根结点的子树 
   */ 
  int judgeChildTree(struct btree *ahead, int numa, struct btree *bhead, int numb) 
  { 
    int flag = 0; 
   
    if (numa != -1 && numb != -1) { 
      if (ahead[numa].value == bhead[numb].value) 
        flag = doesTree1HasTree2(ahead, numa, bhead, numb); 
   
      if (! flag && ahead[numa].lchild != -1) 
        flag = judgeChildTree(ahead, ahead[numa].lchild, bhead, numb); 
   
      if (! flag && ahead[numa].rchild != -1) 
        flag = judgeChildTree(ahead, ahead[numa].rchild, bhead, numb); 
    } 
   
    return flag; 
  } 

  /** 
   * 第二步判断，判断A树是否有B树的子结构 
   */ 
  int doesTree1HasTree2(struct btree *ahead, int numa, struct btree *bhead, int numb) 
  { 
    if (numb == -1)  
      return 1; 
    if (numa == -1) 
      return 0; 
   
    if (ahead[numa].value != bhead[numb].value) 
      return 0; 
   
    return (doesTree1HasTree2(ahead, ahead[numa].lchild, bhead, bhead[numb].lchild) && 
      doesTree1HasTree2(ahead, ahead[numa].rchild, bhead, bhead[numb].rchild)); 
  } 

 #include <stdio.h> 
  #include <stdlib.h> 
   
  // 二叉树结点定义 
  struct btree 
  { 
    int value; 
    int lchild, rchild; 
  }; 
   
  // A树和B树的最多结点数 
  int n, m; 
   
  /** 
   * 第二步判断，判断A树是否有B树的子结构 
   */ 
  int doesTree1HasTree2(struct btree *ahead, int numa, struct btree *bhead, int numb) 
  { 
    if (numb == -1)  
      return 1; 
    if (numa == -1) 
      return 0; 
   
    if (ahead[numa].value != bhead[numb].value) 
      return 0; 
   
    return (doesTree1HasTree2(ahead, ahead[numa].lchild, bhead, bhead[numb].lchild) && 
      doesTree1HasTree2(ahead, ahead[numa].rchild, bhead, bhead[numb].rchild)); 
  } 
   
  /** 
   * 第一步判断，遍历A树查找是否有等于B树根结点的子树 
   */ 
  int judgeChildTree(struct btree *ahead, int numa, struct btree *bhead, int numb) 
  { 
    int flag = 0; 
   
    if (numa != -1 && numb != -1) { 
      if (ahead[numa].value == bhead[numb].value) 
        flag = doesTree1HasTree2(ahead, numa, bhead, numb); 
   
      if (! flag && ahead[numa].lchild != -1) 
        flag = judgeChildTree(ahead, ahead[numa].lchild, bhead, numb); 
   
      if (! flag && ahead[numa].rchild != -1) 
        flag = judgeChildTree(ahead, ahead[numa].rchild, bhead, numb); 
    } 
   
    return flag; 
  } 
   
  int main(void) 
  { 
    int i, data, count, left, right, flag; 
    struct btree *ahead, *bhead; 
   
    while (scanf("%d %d", &n, &m) != EOF) { 
      // 获取A树的节点值 
      ahead = (struct btree *)malloc(sizeof(struct btree) * n); 
      for (i = 0; i < n; i ++) { 
        scanf("%d", &data); 
        ahead[i].value = data; 
        ahead[i].lchild = ahead[i].rchild = -1; 
      } 
   
      for (i = 0; i < n; i ++) { 
        scanf("%d", &count); 
        if (count == 0) { 
          continue; 
        } else { 
          if (count == 1) { 
            scanf("%d", &left); 
            ahead[i].lchild = left - 1; 
          } else { 
            scanf("%d %d", &left, &right); 
            ahead[i].lchild = left - 1; 
            ahead[i].rchild = right - 1; 
          } 
        } 
      } 
   
      // 获取B树的节点值 
      bhead = (struct btree *)malloc(sizeof(struct btree) * m); 
      for (i = 0; i < m; i ++) { 
        scanf("%d", &data); 
        bhead[i].value = data; 
        bhead[i].lchild = bhead[i].rchild = -1; 
      } 
   
      for (i = 0; i < m; i ++) { 
        scanf("%d", &count); 
        if (count == 0) { 
          continue; 
        } else { 
          if (count == 1) { 
            scanf("%d", &left); 
            bhead[i].lchild = left - 1; 
          } else { 
            scanf("%d %d", &left, &right); 
            bhead[i].lchild = left - 1; 
            bhead[i].rchild = right - 1; 
          } 
        } 
      } 
   
      // 判断B树是否为A的子树 
      if (n == 0 || m == 0) { 
        printf("NO\n"); 
        continue; 
      } 
   
      flag = judgeChildTree(ahead, 0, bhead, 0); 
      if (flag) 
        printf("YES\n"); 
      else 
        printf("NO\n"); 
   
      free(ahead); 
      free(bhead); 
    } 
   
    return 0; 
  }