*数论（1）C(n-1,m-1)

Kyoya and Colored Balls

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Kyoya Ootori has a bag with n colored balls that are colored with k different colors. The colors are labeled from 1 to k. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color i before drawing the last ball of color i + 1 for all i from 1 to k - 1. Now he wonders how many different ways this can happen.

//最后一个color[i]必须在最后一个color[i+1]之前;

Input

The first line of input will have one integer k (1 ≤ k ≤ 1000) the number of colors.

Then, k lines will follow. The i-th line will contain c_i, the number of balls of the i-th color (1 ≤ c_i ≤ 1000).

The total number of balls doesn't exceed 1000.

Output

A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1 000 000 007.

Sample test(s)

input

3
2
2
1

output

3

input

4
1
2
3
4

output

1680

Note

In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are:

1 2 1 2 3
1 1 2 2 3
2 1 1 2 3

 

#include<cstdio>
#define ll __int64
const int MOD = 1e9+7;
using namespace std;
ll C[1005][1005];
int a[1005];
int main(){
    C[0][0] = 1;
    for(int i = 1; i < 1005; i++){
        C[0][i] = 1;
        C[i][i] = 1;
        for(int j = 1; j < i; j++)
            C[j][i] = (C[j-1][i-1] + C[j][i-1])%MOD;
    }
    int n;
    scanf("%d",&n);
    for(int i = 0; i < n; i++)
        scanf("%d",&a[i]);
    int cnt = 0;
    int ans = 1;
    for(int i = 1; i <= n; i++){
        ans = (ans*C[a[i-1]-1][cnt + a[i-1] - 1])%MOD;
        cnt += a[i-1];
    }
    printf("%d\n",ans);
    return 0;
}