codeforces 338D GCD Table



D. GCD Table

time limit per test

memory limit per test

input

output

Consider a table G of size n × m such that G(i, j) = GCD(i, j) for all1 ≤ i ≤ n, 1 ≤ j ≤ m.GCD(a, b) is the greatest common divisor of numbersa andb.

You have a sequence of positive integer numbers a₁, a₂, ..., a_k. We say that this sequence occurs in table G if it coincides with consecutive elements in some row, starting from some position. More formally, such numbers1 ≤ i ≤ n and1 ≤ j ≤ m - k + 1 should exist thatG(i, j + l - 1) = a_l for all1 ≤ l ≤ k.

Determine if the sequence a occurs in tableG.

Input

The first line contains three space-separated integers n, m and k (1 ≤ n, m ≤ 10¹²;1 ≤ k ≤ 10000). The second line containsk space-separated integers a₁, a₂, ..., a_k (1 ≤ a_i ≤ 10¹²).

Output

Print a single word "YES", if the given sequence occurs in tableG, otherwise print "NO".

Examples

Input

100 100 5 5 2 1 2 1

Output

YES

Input

100 8 5 5 2 1 2 1

Output

NO

Input

100 100 7 1 2 3 4 5 6 7

Output

NO

Note

Sample 1. The tenth row of table G starts from sequence {1, 2, 1, 2, 5, 2, 1, 2, 1, 10}. As you can see, elements from fifth to ninth coincide with sequencea.

Sample 2. This time the width of table G equals 8. Sequencea

【分析】

扩展中国剩余定理。

¡ 转化问题：

¡ gcd(x,y)=a0, gcd(x,y+1)=a1, gcd(x,y+2)=a2……

¡ → y=-k(modak) 是原式有解的必要条件

¡ 代入 验证，若原始条件成立则 x,y 由中国 剩 余 定理可得 x=lcm({ ak }) ， y 有唯一解

¡ 即 为答案，否则问题无解。

¡ 易知 x 最小为 lcm({ak}) ，在此时根据中国剩余定理 y 有唯一解，但是这个解只保证了 ak|x,y ，并没有保证 ak=gcd(x,y) ，由于随着 x,y 增大若干倍会更加不满足，所以此时是最有可能有解的情况，代入验证即可。

【代码】

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(ll i=j;i<=k;i++)
using namespace std;
const int mxn=10005;
ll n,m,k;
ll a[mxn],b[mxn],w[mxn];
inline ll gcd(ll x,ll y)
{
  return x%y==0?y:gcd(y,x%y);
}
inline void exgcd(ll aa,ll bb,ll &x,ll &y)
{
  if(!bb)
  {
    x=1,y=0;
    return;
  }
  exgcd(bb,aa%bb,y,x);
  y-=(aa/bb)*x;
}
inline ll inv(ll aa,ll bb)
{
  ll x,y;
  exgcd(aa,bb,x,y);
  return (x%bb+bb)%bb;
}
inline ll mul(ll aa,ll bb,ll mod)  //快速乘 
{
  ll ans=0;
  if(bb<0) aa=-aa,bb=-bb;
  while(bb)
  {
    if(bb&1) ans=(ans+aa)%mod;
    bb>>=1;aa=(aa+aa)%mod;
  }
  return ans;
}
inline bool judge()
{
  ll lcm=1,a1,a2,b1,b2,t,tmp=a[1];
  fo(i,1,k)
  {
    b[i]=1-i;
    t=gcd(lcm,a[i]); 
    lcm=lcm/t*a[i];
    if(lcm>n) return 0;
  }
  fo(i,2,k)
  {
    a1=a[i-1],a2=a[i],b1=b[i-1],b2=b[i];
    t=gcd(a1,a2);
    if((b2-b1)%t!=0) return 0;
    a[i]=a1/t*a2;
    b[i]=inv(a1/t,a2/t);
    b[i]=b[i];
    b[i]=mul(b[i],(b2-b1)/t,a2/t);
    b[i]=mul(b[i],a1,a[i]);
    b[i]=(b[i]+b1)%a[i];
    b[i]=(b[i]%a[i]+a[i])%a[i];
  }
  ll y=b[k];
  if(!y) y+=a[k];
  if(y+k-1>m) return 0;
  fo(j,y,y+k-1)
    if(gcd(lcm,j)!=w[j-y+1]) return 0;
  return 1;
}
int main()
{
  scanf("%I64d%I64d%I64d",&n,&m,&k);
  fo(i,1,k) scanf("%I64d",&a[i]),w[i]=a[i];
  if(judge()) printf("YES\n");
  else printf("NO\n");
  return 0;
}