等差数列可以用 \(gcd\) 来维护，这很显然。 但是本题有一个限制是取模，所以 \(gcd\) 直接寄了，换一个做法（类似于随机化的想法，就是说 \(k\) 次方的和相等，这样可以保证与顺序无关）。
推一下式子。
等式首项 $$a = \frac{\Sigma_{i=l}^r N_i-\frac{len\times (len-1) *d}{2}}{len}$$
知道首项之后就要求
二项式定理可以推得
\[\Sigma(a+di)^k=\Sigma \binom{k}{j} d ^j a ^{k - j} \Sigma_{i=0}^{len - 1} i ^ j \]Tips:
如果顺序无关，直接判断 \(k\) 次方和是否相等就行，正确性可以保证。
#include<bits/stdc++.h>
#define RG register
#define LL long long
#define U(x, y, z) for(RG int x = y; x <= z; ++x)
#define D(x, y, z) for(RG int x = y; x >= z; --x)
#define update(x, y) (x = x + y >= mod ? x + y - mod : x + y)
using namespace std;
const int mod = 1e9 + 7;
namespace FastIO {
#define il inline
const int iL = 1 << 25;
char ibuf[iL], *iS = ibuf + iL, *iT = ibuf + iL;
#define GC() (iS == iT) ? \
(iT = (iS = ibuf) + fread(ibuf, 1, iL, stdin), (iS == iT) ? EOF : *iS++) : *iS++
void read(){}
template<typename _Tp, typename... _Tps>
void read(_Tp &x, _Tps &...Ar) {
x = 0; char ch = GC(); bool flg = 0;
for (; !isdigit(ch); ch = GC()) flg |= (ch == '-');
for (; isdigit(ch); ch = GC()) x = (x << 1) + (x << 3) + (ch ^ 48);
if (flg) x = -x;
read(Ar...);
}
char Out[iL], *iter = Out;
#define Flush() fwrite(Out, 1, iter - Out, stdout); iter = Out
template <class T>il void write(T x, char LastChar = '\n') {
int c[35], len = 0;
if (x < 0) {*iter++ = '-'; x = -x;}
do {c[++len] = x % 10; x /= 10;} while (x);
while (len) *iter++ = c[len--] + '0';
*iter++ = LastChar; Flush();
}
template <typename T> inline void writeln(T n){write(n, '\n');}
template <typename T> inline void writesp(T n){write(n, ' ');}
inline char Getchar(){ char ch; for (ch = GC(); !isalpha(ch); ch = GC()); return ch;}
inline void readstr(string &s) { s = ""; static char c = GC(); while (isspace(c)) c = GC(); while (!isspace(c)) s = s + c, c = GC();}
}
using namespace FastIO;
struct modint{
int x;
modint(int o=0){x=o;}
modint &operator = (int o){return x=o,*this;}
modint &operator +=(modint o){return x=x+o.x>=mod?x+o.x-mod:x+o.x,*this;}
modint &operator -=(modint o){return x=x-o.x<0?x-o.x+mod:x-o.x,*this;}
modint &operator *=(modint o){return x=1ll*x*o.x%mod,*this;}
modint &operator ^=(int b){
if(b<0)return x=0,*this;
b%=mod-1;
modint a=*this,c=1;
for(;b;b>>=1,a*=a)if(b&1)c*=a;
return x=c.x,*this;
}
modint &operator /=(modint o){return *this *=o^=mod-2;}
modint &operator +=(int o){return x=x+o>=mod?x+o-mod:x+o,*this;}
modint &operator -=(int o){return x=x-o<0?x-o+mod:x-o,*this;}
modint &operator *=(int o){return x=1ll*x*o%mod,*this;}
modint &operator /=(int o){return *this *= ((modint(o))^=mod-2);}
template<class I>friend modint operator +(modint a,I b){return a+=b;}
template<class I>friend modint operator -(modint a,I b){return a-=b;}
template<class I>friend modint operator *(modint a,I b){return a*=b;}
template<class I>friend modint operator /(modint a,I b){return a/=b;}
friend modint operator ^(modint a,int b){return a^=b;}
friend bool operator ==(modint a,int b){return a.x==b;}
friend bool operator !=(modint a,modint b){return a.x!=b.x;}
bool operator ! () {return !x;}
modint operator - () {return x?mod-x:0;}
};
template <typename T> inline void chkmin(T &x, T y){x = x < y ? x : y;}
template <typename T> inline void chkmax(T &x, T y){x = x > y ? x : y;}
template <typename T> inline T Min(T x, T y){return x < y ? x : y;}
template <typename T> inline T Max(T x, T y){return x > y ? x : y;}
inline void FO(string s){freopen((s + ".in").c_str(), "r", stdin); freopen((s + ".out").c_str(), "w", stdout);}
const int N = 2e5 + 10, M = 12;
int n, T;
modint a[N], b[N], s[N][M], inv[N], pw[N][M], spw[N][M], C[M][M];
modint Qpow(modint a, LL b) {
modint res = 1;
// cerr << "Chk " << a.x << " " << b << " ";
for (; b; b >>= 1) {
if (b & 1) res *= a;
a *= a;
}
// cerr << res.x << "\n";
return res;
}
inline void init() {
U(i, 1, n) b[i] = 1;
U(j, 1, 10) U(i, 1, n) {
b[i] *= a[i];
s[i][j] = s[i - 1][j] + b[i];
}
inv[1] = 1;
U(i, 2, n) inv[i] = inv[mod % i] * (mod - mod / i);
U(i, 0, n) pw[i][0] = 1;
U(i, 1, n) U(j, 1, 10) pw[i][j] = pw[i][j - 1] * i;
U(j, 0, 10) spw[0][j] = pw[0][j];
U(j, 0, 10) U(i, 1, n) spw[i][j] = spw[i - 1][j] + pw[i][j];
U(i, 0, 10) C[i][0] = 1;
U(i, 1, 10) U(j, 1, i) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
modint Dcalc(int k,modint A,modint d,int r){
modint tot=0;
U(i, 0, k){
tot=(tot+C[k][i]*Qpow(A,k-i)*Qpow(d,i)*spw[r][i]);
// cerr << C[k][i].x << " " << A.x << " " << Qpow(A, k - i).x << " " << Qpow(d, i).x << " " << spw[r][i].x << "\n";
}
return tot;
}
bool calc(modint A,modint d,int l,int r){
int k;
k=2;if( Dcalc(k,A,d,r-l)!=( (s[r][k]-s[l-1][k]) ) )return 0;
k=3;if( Dcalc(k,A,d,r-l)!=( (s[r][k]-s[l-1][k]) ) )return 0;
k=6;if( Dcalc(k,A,d,r-l)!=( (s[r][k]-s[l-1][k]) ) )return 0;
k=7;if( Dcalc(k,A,d,r-l)!=( (s[r][k]-s[l-1][k]) ) )return 0;
k=8;if( Dcalc(k,A,d,r-l)!=( (s[r][k]-s[l-1][k]) ) )return 0;
k=9;if( Dcalc(k,A,d,r-l)!=( (s[r][k]-s[l-1][k]) ) )return 0;
return 1;
}
int main(){
//FO("");
read(n, T);
U(i, 1, n) read(a[i].x);
init();
while (T--) {
int l, r; modint d;
read(l, r, d.x);
modint A = (s[r][1] - s[l - 1][1] - d * (r - l) * (r - l + 1) * inv[2]) * inv[r - l + 1];
puts(calc(A, d, l, r) ? "Yes" : "No");
}
return 0;
}