#include #include #define rep(i,n) for(int i=0;i vi; typedef vector vl; typedef vector> vvi; typedef vector> vvl; typedef long double ld; typedef pair P; ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;} template ostream& operator<<(ostream& os, const static_modint& a) {os << a.val(); return os;} template ostream& operator<<(ostream& os, const dynamic_modint& a) {os << a.val(); return os;} template istream& operator>>(istream& is, vector& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;} template ostream& operator<<(ostream& os, const pair& p){os << p.first << ' ' << p.second; return os;} template ostream& operator<<(ostream& os, const vector& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const vector>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;} template ostream& operator<<(ostream& os, const set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const unordered_set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const atcoder::segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const atcoder::lazy_segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template void chmin(T& a, T b){a = min(a, b);} template void chmax(T& a, T b){a = max(a, b);} int solve(){ int n, k, x; cin >> n >> k >> x; vector a(n); cin >> a; vector d(n + 1); for(int i = 1; i < n; i++) d[i] = a[i] - a[i - 1]; string ans = "Yes\n"; rep(i, k){ bool front; bool back; if(i == 0) front = true; if(n % k == i) back = true; vector vec; for(int j = i; j <= n; j+= k){ if(j != 0 and j != n) vec.push_back(d[j]); } // cout << vec; long long sum = 0; for(long long val : vec) sum += val; long long g = abs(d[i]); for(long long val : vec) g = gcd(g, abs(val)); if(g % x != 0) ans = "No\n"; if(!front and !back) if(sum != 0) ans = "No\n"; long long first = d[i]; long long last = d[n / k * k + n % k]; if(!front and first > 0) ans = "No\n"; if(!back and last < 0) ans = "No\n"; // cout << i << ' ' << ans; } cout << ans; return 0; } int main(){ int t; cin >> t; rep(_, t) solve(); return 0; }