Codeforces 964C Alternating Sum

You are given two integers $a$ and $b$. Moreover, you are given a sequence $s_0,s_1,\dots ,s_n$. All values in $s$ are integers $1$ or $-1$. It's known that sequence is $k$-periodic and $k$ divides $n+1$. In other words, for each $k\le i\le n$ it's satisfied that $s_i=s_{i-k}$.Find out the non-negative remainder of division of $\sum _{i=0}^n{s}_i{a}^{n-i}{b}^i$ by ${10}^9+9$.Note that the modulo is unusual!

Input

The first line contains four integers $n,a,b$ and $k$ $(1\le n\le {10}^9,1\le a,b\le {10}^9,1\le k\le {10}^5)$.

The second line contains a sequence of length $k$ consisting of characters '+' and '-'. If the $i$ -th character (0-indexed) is'+', then $s_i=1$, otherwise $s_i=-1$.

Note that only the first $k$members of the sequence are given, the rest can be obtained using the periodicity property.

Output

Output a single integer — value of given expression modulo ${10}^9+9$.

Examples

Input

2 2 3 3
+-+

Output

7

Input

4 1 5 1
-

Output

999999228

Note

In the first example:

$\left(\sum _{i=0}^n{s}_i{a}^{n-i}{b}^i\right)$

= $2^2{3}^0-2^1{3}^1+2^0{3}^2$

= 7

In the second example:

$\left(\sum _{i=0}^n{s}_i{a}^{n-i}{b}^i\right)=-1^4{5}^0-1^3{5}^1-1^2{5}^2-1^1{5}^3-1^0{5}^4=-781\equiv 999999228(\mathrm{mod}{10}^9+9）$

$题意：$

$给你一个公式，给你一个k个字符，表示前k个的加减规则，然后就是让你求（n+1)/k\; 这段长度中的值，对1e9+9取余。$

$解题思路：$

$我们可以很快的从题目中看到这是一个等比数列，求前（n+1)/k项的和，比例系数为（b/a)^k\; 但是我们不能忘记了这个比例系数为1的特殊情况。然后就是用快速幂加速，用扩展GCD求逆元。$

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define  ll long long
using namespace std;
const long long mo=1e9+9;
const int maxn=1e5+10;
ll n,a,b,k;
char ch[maxn];
ll find(ll a,ll t){
	ll res=1;
	while(t){
		if(t&1) res=res*a%mo;
		a=(a*a)%mo;
		t/=2;
	}
	return (res+mo)%mo;
}
ll ex_gcd(ll a,ll b,ll &x,ll &y){
	if(!b){
		x=1,y=0;return a;
	}
	ll te=ex_gcd(b,a%b,x,y);
	ll t=x;
	x=y;
	y=t-(a/b)*x;
	return te;
}
ll NY(ll nu){
	ll x,y;  
	ex_gcd(nu,mo,x,y);
	return (x+mo)%mo;
}
int main(){
	int i,j;
	scanf("%I64d%I64d%I64d%I64d",&n,&a,&b,&k);
	scanf("%s",ch);
	ll ans=0,m=(n+1)/k;
	for(i=0;i<k;i++){
		ll t1=find(a,n-i),t2=find(b,i);
		ll temp=t1*t2%mo;
		if(ch[i]=='+')
			ans+=temp;
		else
			ans-=temp;
		ans+=mo;
		ans%=mo;
	}
	ll q=find(NY(a),k)*find(b,k)%mo;
	if(q==1){
		ans=ans*m%mo;
	}
	else{
		long long t1=NY(q-1)%mo;
		ans=(ans*(find(q,m)-1)%mo)*t1%mo;
		ans%=mo;
	}
	ans=(ans+mo)%mo;
	printf("%I64d\n",ans);
	return 0;
}

.