D. Colored Balls

There are balls of $n$ different colors; the number of balls of the $i$-th color is $a_i$.

The balls can be combined into groups. Each group should contain at most $2$ balls, and no more than $1$ ball of each color.

Consider all $2^n$ sets of colors. For a set of colors, let's denote its value as the minimum number of groups the balls of those colors can be distributed into. For example, if there are three colors with $3$, $1$ and $7$ balls respectively, they can be combined into $7$ groups (and not less than $7$), so the value of that set of colors is $7$.

Your task is to calculate the sum of values over all $2^n$ possible sets of colors. Since the answer may be too large, print it modulo $998\,244\,353$.

Input

The first line contains a single integer $n$ ($1 \le n \le 5000$) — the number of colors.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 5000$) — the number of balls of the $i$-th color.

Additional constraint on input: the total number of balls doesn't exceed $5000$.

Output

Print a single integer — the sum of values of all $2^n$ sets of colors, taken modulo $998\,244\,353$.

Examples

input

3
1 1 2

output

11

input

1
5

output

5

input

4
1 3 3 7

output

76

Note

Consider the first example. There are $8$ sets of colors:

• for the empty set, its value is $0$;
• for the set $\{1\}$, its value is $1$;
• for the set $\{2\}$, its value is $1$;
• for the set $\{3\}$, its value is $2$;
• for the set $\{1,2\}$, its value is $1$;
• for the set $\{1,3\}$, its value is $2$;
• for the set $\{2,3\}$, its value is $2$;
• for the set $\{1,2,3\}$, its value is $2$.

So, the sum of values over all $2^n$ sets of colors is $11$.

解题思路

  题目给出的组合规则其实就是摩尔投票。即确定颜色的方案后，假设球的总数为 $m$，所有颜色中数量最多的球有 $x$ 个，则分组数量的最小值分两种情况：

• 如果 $x \leq m-x$，则至少分成 $\left\lceil \frac{n}{2} \right\rceil$ 组。
• 如果 $x > m-x$，则至少分成 $x$ 组。

  所以我们可以对数组 $a$ 升序排序，固定数量最多的球 $a_i$，用 $f(i-1,j)$ 来表示所有不超过 $a_i$ 的球（即 $a_1 \sim a_{i-1}$）组成总数为 $j$ 的方案数量。那么以 $a_i$ 作为数量最多的球的所有方案中，贡献的答案是（其中 $m = \sum{a_i}$）：

$$\sum\limits_{j=0}^{a_i}{\left\lceil \frac{j+a_i}{2} \right\rceil \cdot f(i-1,j)} + \sum\limits_{j=a_i+1}^{m}{a_i \cdot f(i-1,j)}$$

  其中 $f(i, j)$ 根据是否选择 $a_i$ 进行状态划分（01 背包），转移方程为 $f(i,j) = f(i-1,j) + f(i-1,j-a_i)$。

  AC 代码如下，时间复杂度为 $O(n \cdot \sum{a_i})$：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 5005, mod = 998244353;

int a[N];
int f[N][N];

int main() {
    int n, m = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", a + i);
        m += a[i];
    }
    sort(a + 1, a + n + 1);
    int ret = 0;
    f[0][0] = 1;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            if (j >= a[i]) ret = (ret + (j + a[i] + 1ll) / 2 * f[i - 1][j]) % mod;
            else ret = (ret + 1ll * a[i] * f[i - 1][j]) % mod;
            f[i][j] = f[i - 1][j];
            if (j >= a[i]) f[i][j] = (f[i][j] + f[i - 1][j - a[i]]) % mod;
        }
    }
    printf("%d", ret);
    
    return 0;
}

参考资料

  Educational Codeforces Round 164 (Rated for Div. 2) A~E - 知乎：https://zhuanlan.zhihu.com/p/692227075