import heapq

def main():
    import sys
    input = sys.stdin.read
    data = input().split()
    idx = 0
    N = int(data[idx])
    idx += 1
    M = int(data[idx])
    idx += 1
    m = list(map(int, data[idx:idx+M]))
    idx += M
    m = [0] + m  # 1-based indexing for m[1..M]
    
    INF = float('inf')
    # d[u][i] is the minimal cost to reach node u with last operation i
    d = [[INF] * (M + int(1)) for _ in range(N + int(1))]  # u:1..N, i:1..M
    
    heap = []
    # Initialize: start at node 1, using any operation i (cost 0)
    for i in range(1, M + 1):
        d[1][i] = 0
        heapq.heappush(heap, (0, 1, i))
    
    while heap:
        dist, u, i = heapq.heappop(heap)
        if u == N:
            print(dist)
            return
        if dist > d[u][i]:
            continue
        # Try all operations j
        for j in range(1, M + 1):
            mj = m[j]
            v1 = u + mj
            v2 = u - mj
            vs = []
            if 1 <= v1 <= N:
                vs.append(v1)
            if 1 <= v2 <= N:
                vs.append(v2)
            # Avoid duplicate v1 and v2
            vs = list(set(vs))
            for v in vs:
                new_cost = dist + abs(2 * mj) + abs(j - i)
                if new_cost < d[v][j]:
                    d[v][j] = new_cost

省流题面：一张无向图中有 $N$ 个点，给定 $M$ 个操作（以下用 $m_1,m_2,...,m_n$ 表示）， 如果来到当前节点使用的操作为 $m_i$ ，下一步可以花费 $ \left| 2 \times m_j \right| + \left| j-i \right| $ 的代价移动 $ m_j $ 个单位，也可以花费 $ \left| 2 \times m_i \right| $ 的代价移动 $ m_i $ 个单位，求从 $ 1 $ 号点到 $ N $ 号点的最少代价。