Problem Description

You are given one integer number n. Find three distinct integers a,b,c such that 2≤a,b,c and a?b?c=n or say that it is impossible to do it.

If there are several answers, you can print any.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1≤t≤100) — the number of test cases.

The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2≤n≤109).

Output

For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a?b?c for some distinct integers a,b,c such that 2≤a,b,c.

Otherwise, print "YES" and any possible such representation.

Examples

Input

5
64
32
97
2
12345

Output

YES
2 4 8
NO
NO
NO
YES
3 5 823

題意：t 組數據，每組給出一個數 n，問 n 是否能分解成 a*b*c 的形式，要求 a、b、c 大于等于 2 并且不相等，若存在，輸出任意一個方案

思路：

將 n 分解成 a*b*c 的形式，那么 a、b、c 這三個數一定是 n 的因子，因此首先將 n 進行因子分解，并統計其素因子個數，然后判斷是否有三個不一樣的因子

當 n 存在一個素因子 x 時：最簡單的方案是 x * (x*x) * (x*x*x)，也即 n 中至少存在 6 個以上的 x 時，才存在合法方案

當 n 存在兩個素因子 x1、x2 時：最簡單的方案是 (x1*x2)*x1*x2、(x1*x1)*x1*x2、(x2*x2)*x1*x2，也即 n 中的 x1、x2 個數和至少達到 4 個時，才存在合法方案

當 n 存在三個及以上的素因子時，一定存在合法方案

而輸出合法方案是一個問題，可以發現，如果有了 a、b，那么 c 就不需要進行暴力枚舉，直接利用 c=n/(a*b) 即可計算得出

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long #define Pair pair<int,int> LL quickPow(LL a,LL b){ LL res=1; while(b){if(b&1)res*=a; a*=a; b>>=1;} return res; } LL multMod(LL a,LL b,LL mod){ a%=mod; b%=mod; LL res=0; while(b){if(b&1)res=(res+a)%mod; a=(a<<=1)%mod; b>>=1; } return res%mod;} LL quickMultPowMod(LL a, LL b,LL mod){ LL res=1,k=a; while(b){if((b&1))res=multMod(res,k,mod)%mod; k=multMod(k,k,mod)%mod; b>>=1;} return res%mod;} LL quickPowMod(LL a,LL b,LL mod){ LL res=1; while(b){if(b&1)res=(a*res)%mod; a=(a*a)%mod; b>>=1; } return res; } LL getInv(LL a,LL mod){ return quickPowMod(a,mod-2,mod); } LL GCD(LL x,LL y){ return !y?x:GCD(y,x%y); } LL LCM(LL x,LL y){ return x/GCD(x,y)*y; } const double EPS = 1E-6; const int MOD = 1000000000+7; const int N = 1000+5; const int dx[] = {0,0,-1,1,1,-1,1,1}; const int dy[] = {1,-1,0,0,-1,1,-1,1}; using namespace std;map<LL, int> num; //記錄素因子個數 LL factor[N], cntF; //記錄素因子 void getFactor(LL n) {cntF = 0;num.clear();for (LL i = 2; i <= n / i; i++) {if (n % i == 0) //記錄素因子factor[cntF++] = i;while (n % i == 0) {n /= i;num[i]++; //記錄素因子個數}}if (n > 1) {factor[cntF++] = n;num[n]++;} } int main() {int t;scanf("%d", &t);while (t--) {LL n;scanf("%lld", &n);getFactor(n);if (cntF == 1) {if (num[factor[0]] >= 6) {LL a = factor[0];LL b = factor[0] * factor[0];LL c = n / (a * b);printf("YES\n");printf("%lld %lld %lld\n", a, b, c);} elseprintf("NO\n");} else if (cntF == 2) {if (num[factor[0]] + num[factor[1]] >= 4) {LL a = factor[0];LL b = factor[1];LL c = n / (a * b);printf("YES\n");printf("%lld %lld %lld\n", a, b, c);} elseprintf("NO\n");} else {LL a = factor[0];LL b = factor[1];LL c = n / (a * b);printf("YES\n");printf("%lld %lld %lld\n", a, b, c);}}return 0; }

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