# 2次元dpでは間に合わないか
# dp[i番目][j0 for -1, j1 for +1]でのf(s)最小値

N, A, B = map(int, input().split())

dp = [[0]*2 for i in range(N)]

for i in range(N):
    if i == 0:
        dp[0][0] = -B*(-1)
        dp[0][1] = -B*(+1)
    else:
        dp[i][0] = min(dp[i-1][0]+A-B*(-1), dp[i-1][1]-A-B*(-1))
        dp[i][1] = min(dp[i-1][0]-A-B*(+1), dp[i-1][1]+A-B*(+1))
    #print(dp[i])
    
ans = min(dp[N-1])
print(ans)