Codeforces 554C. Kyoya and Colored Balls【组合数 逆元】

C. 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 ibefore 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.

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 ci, the number of balls of the i-th color (1 ≤ ci ≤ 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.

Examples

input

3221

output

3

input

41234

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 31 1 2 2 32 1 1 2 3

题意是，第i 中颜色的最后一个球后面不能出现前 i-1中颜色的球，由于数据范围，用到逆元

//首先考虑最后一种颜色a个，总共s个，必有一个球是放在最后的C（a-1,s-1)之后再如此考虑倒数第二种b个，C（b-1,s-a-1)累乘#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#define maxn 1000010#define MOD 1000000007using namespace std;int num[1010];__int64 r[maxn];//开始数组开小了void mul(){    r[0]=1;    for(__int64 i=1;i<=1000000;++i)        r[i]=(i*r[i-1])%MOD;}__int64 pow_mul(__int64 a,__int64 b){    __int64 sum=1;    while(b)    {        if(b&1)            sum=sum*a%MOD;        a=a*a%MOD;        b=b/2;    }    return sum;}int main(){    int k;    mul();    while(~scanf("%d",&k))    {        int sum_num=0;        for(int i=0;i<k;++i)        {            scanf("%d",&num[i]);            sum_num+=num[i];        }        __int64 ans=1;        for(int i=k-1;i>=0;--i)        {            ans=ans*(r[sum_num-1]*(pow_mul((r[num[i]-1]*r[sum_num-num[i]]%MOD),MOD-2)%MOD)%MOD)%MOD;            sum_num-=num[i];//这里变量名弄错找好久        }        printf("%I64d\n",ans);    }    return 0;}