Codeforces Round #419 (Div. 2) (Codeforces 815B) D. Karen and Test 组合数学

D. Karen and Test

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Karen has just arrived at school, and she has a math test today!

The test is about basic addition and subtraction. Unfortunately, the teachers were too busy writing tasks for Codeforces rounds, and had no time to make an actual test. So, they just put one question in the test that is worth all the points.

There are n integers written on a row. Karen must alternately add and subtract each pair of adjacent integers, and write down the sums or differences on the next row. She must repeat this process on the values on the next row, and so on, until only one integer remains. The first operation should be addition.

Note that, if she ended the previous row by adding the integers, she should start the next row by subtracting, and vice versa.

The teachers will simply look at the last integer, and then if it is correct, Karen gets a perfect score, otherwise, she gets a zero for the test.

Karen has studied well for this test, but she is scared that she might make a mistake somewhere and it will cause her final answer to be wrong. If the process is followed, what number can she expect to be written on the last row?

Since this number can be quite large, output only the non-negative remainder after dividing it by 109 + 7.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 200000), the number of numbers written on the first row.

The next line contains n integers. Specifically, the i-th one among these is ai (1 ≤ ai ≤ 109), the i-th number on the first row.

Output

Output a single integer on a line by itself, the number on the final row after performing the process above.

Since this number can be quite large, print only the non-negative remainder after dividing it by 109 + 7.

Examples

input

53 6 9 12 15

output

36

input

43 7 5 2

output

1000000006

Note

In the first test case, the numbers written on the first row are 3, 6, 9, 12 and 15.

Karen performs the operations as follows:

The non-negative remainder after dividing the final number by 109 + 7 is still 36, so this is the correct output.

In the second test case, the numbers written on the first row are 3, 7, 5 and 2.

Karen performs the operations as follows:

The non-negative remainder after dividing the final number by 109 + 7 is 109 + 6, so this is the correct output.

一列n个数，把这些数两两按照+，-，+，- ...的运算计算，每次把n个数合并成n-1个，问最后剩下的一个数mod 1e9+7 是多少。

经过一番严谨的XJB猜测，发现根据n%4的四种结果，可以找到每个数字在最后答案当中贡献的系数的规律。这些系数可以用组合数表示。详细规律如下：（来自CF官方题解）

When , the pattern is: 

When , the pattern is: 

When , the pattern is: 

When , the pattern is: 

接着只要算组合数就可以了。组合数太大，需要计算逆元。

一个关于阶乘逆元的神奇规律：

inv[n!]=inv[(n+1)!]*(n+1)

由此，可以容易地算出大组合数。

#include <cstdio>#include <iostream>#include <string.h>#include <string> #include <map>#include <queue>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <stack>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef long long ll;typedef long double ld;const int maxn=200005,inf=0x3f3f3f3f;  const ll llinf=0x3f3f3f3f3f3f3f3f,mod=1e9+7;   const ld pi=acos(-1.0L);  ll a[maxn],inv[maxn];ll fastpow(ll base,ll index) {ll sum=base,an=1;ll i=index;    while (i) {    if (i%2) an=(an*sum)%mod;    sum*=sum;    sum=sum%mod;    i/=2;    }    return an;}int main() {int n,i,j;ll ans=0,fac=1,g;scanf("%d",&n);for (i=1;i<=n;i++) scanf("%I64d",&a[i]);if (n%2==0) g=(n-2)/2; else g=(n-n%4)/2;for (i=1;i<=g;i++) fac=(fac*((ll)i))%mod;inv[g]=fastpow(fac,mod-2);for (i=g-1;i>=0;i--) {inv[i]=inv[i+1]*(i+1);inv[i]%=mod;}if (n%2==0) {ll t=1;for (i=1;i<=n;i++) {if (i%2==0&&n%4==0) ans-=t*a[i]; else ans+=t*a[i];ans%=mod;if (ans<0) ans+=mod;if (i%2==0) {t=(((fac*inv[i/2])%mod)*inv[g-i/2])%mod;}}} else if (n%4==1) {ll t=1;for (i=1;i<=n;i++) {if (i%2!=0) {ans+=t*a[i];ans%=mod;} else {t=(((fac*inv[i/2])%mod)*inv[g-i/2])%mod;}}} else if (n%4==3) {ll t=1,last=0;for (i=1;i<=n;i++) {if (i%2==0) ans+=2*t*a[i]; else ans+=(t-last)*a[i];ans%=mod;if (ans<0) ans+=mod;if (i%2==0) {last=t;t=(((fac*inv[i/2])%mod)*inv[g-i/2])%mod;}}}printf("%I64d\n",ans);return 0;}